## Roots of 1 + x + x^2/2! + … + x^n/n! (1)

Problem. Given an integer $n \ge 0,$ consider the polynomial $\displaystyle p(x):=\sum_{k=0}^n \frac{x^k}{k!}=1+x+\frac{x^2}{2!} + \cdots + \frac{x^n}{n!}.$
Show that
i) if $n$ is even, $p(x) > 0$ for all $x$ and so $p(x)$ has no real root,
ii) if $n$ is odd, $p(x)$ has exactly one real root and the root is $< -1$ for $n > 1,$
iii) if $n$ is even, $\displaystyle p(x) \ge \frac{1}{n!}$ for all $x.$

Solution. By Taylor’s theorem, for every real number $x,$ there exists $c$ between $0$ and $x$ such that $\displaystyle e^x=p(x)+\frac{e^c}{(n+1)!}x^{n+1}$ and so
$\displaystyle p(x)=e^x - \frac{e^c}{(n+1)!}x^{n+1}. \ \ \ \ \ \ \ \ \ \ \ \ (*)$

i) Clearly $p(x) > 0$ for all $x \ge 0.$ For $x < 0,$ there exists $x < c < 0$ such that $(*)$ holds. Since $x < 0$ and $n+1$ is odd, $x^{n+1} < 0$ and so $\displaystyle p(x)=e^x - \frac{e^c}{(n+1)!}x^{n+1} > 0.$

ii) If $n=1,$ then $p(x)=1+x$ and so $x=-1$ is the root of $p(x).$ Suppose now that $n > 1.$ Since $n-1$ is even, $\displaystyle p'(x)= \sum_{k=0}^{n-1} \frac{x^k}{k!}$  has no real root, by i). Thus $p(x)$ has at most one real root, by Rolle’s theorem. Since $\displaystyle \lim_{x\to-\infty}p(x)=-\infty,$ (because $n$ is odd) and $p(0)=1 > 0, \ p(x)$ has a root in the interval $(-\infty, 0),$ by the intermediate value theorem. So we have proved that $p(x)$ has a unique real root $\alpha$ and $\alpha < 0.$
To prove that $\alpha < -1,$ we only need to show that $p(-1) > 0.$ By $(*),$ there exists $-1 < c < 0$ such that $\displaystyle p(-1)=e^{-1} - \frac{e^c}{(n+1)!}$ and so, since $e^c < 1$ and $n > 2,$ we have $\displaystyle p(-1) > e^{-1}-\frac{1}{6} > 0.$

iii) If $n=0,$ then $p(x)=1$ and there’s nothing to prove. Suppose now that $n > 0.$ By ii), $p'(x)$ has a unique real root $\alpha$ and $\alpha \le -1.$ Since, by i), $p''(x) > 0, \ p(x)$ attains its absolute minimum at $x = \alpha.$ Thus $\displaystyle p(x) \ge p(\alpha)=\frac{\alpha^n}{n!} \ge \frac{1}{n!}. \ \Box$

In part (2) of this post, we are going to dig a little deeper and get better results.

## The Stolz–Cesàro lemma

Problem (The Stolz–Cesàro lemma). Let $\{a_n\}, \{b_n\}$ be two sequences of real numbers. Suppose that
i) $b_n > 0$ for all $n$ and $\{b_n\}$ is strictly increasing,
ii) $\displaystyle \lim_{n\to\infty}b_n=\infty$ and $\displaystyle \lim_{n\to\infty} \frac{a_n-a_{n-1}}{b_n-b_{n-1}}=\ell \in [-\infty,+\infty].$
Show that $\displaystyle \lim_{n\to\infty} \frac{a_n}{b_n}=\ell.$

Solution. I’ll assume that $\ell$ is a finite number; the proof for $\ell = \pm \infty$ is similar (Exercise 1).
Let $\epsilon > 0.$ Then, by the definition of limit, there exists an integer $N$ such that if $n > N,$ then
$\displaystyle \ell - \epsilon < \frac{a_n-a_{n-1}}{b_n-b_{n-1}} < \ell + \epsilon. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$
Since $\{b_n\}$ is strictly increasing, $b_n-b_{n-1} > 0$ and so $(1)$ gives $\displaystyle (\ell - \epsilon)(b_n-b_{n-1}) < a_n-a_{n-1} < (\ell + \epsilon)(b_n-b_{n-1}),$
for all $n > N.$ Therefore
$\displaystyle (\ell-\epsilon)\sum_{k=1}^{n-N} (b_{N+k}-b_{N+k-1})<\sum_{k=1}^{n-N}(a_{N+k}-a_{N+k-1}) < (\ell + \epsilon)\sum_{k=1}^{n-N} (b_{N+k}-b_{N+k-1})$
Thus, since $\displaystyle \sum_{k=1}^{n-N} (b_{N+k}-b_{N+k-1})=b_n-b_N$ and $\displaystyle \sum_{k=1}^{n-N}(a_{N+k}-a_{N+k-1})=a_n-a_N,$ we get
$\displaystyle (\ell-\epsilon)(b_n-b_N) < a_n-a_N < (\ell+\epsilon)(b_n-b_N),$
which gives us
$\displaystyle (\ell-\epsilon)\left(1-\frac{b_N}{b_n}\right)+\frac{a_N}{b_n} < \frac{a_n}{b_n} < (\ell + \epsilon)\left(1-\frac{b_N}{b_n}\right)+\frac{a_N}{b_n}. \ \ \ \ \ \ \ \ \ \ \ \ (2)$
Since $\displaystyle \lim_{n\to\infty}b_n=\infty,$ we have $\displaystyle \lim_{n\to\infty} \frac{a_N}{b_n}=\lim_{n\to\infty}\frac{b_N}{b_n}=0$ and so
$\displaystyle \lim_{n\to\infty} (\ell \pm \epsilon)\left(1-\frac{b_N}{b_n}\right)+\frac{a_N}{b_n}=\ell \pm \epsilon.$
Therefore if $n$ is large enough, we will have
$\displaystyle (\ell - \epsilon)\left(1-\frac{b_N}{b_n}\right)+\frac{a_N}{b_n} > \ell -2\epsilon, \ \ (\ell + \epsilon)\left(1-\frac{b_N}{b_n}\right)+\frac{a_N}{b_n} < \ell+2\epsilon$
and then, by $(2),$ $\displaystyle \ell-2\epsilon < \frac{a_n}{b_n} < \ell + 2\epsilon,$ proving that $\displaystyle \lim_{n\to\infty} \frac{a_n}{b_n}=\ell. \ \Box$

Example 1. Let $\{u_n\}, \{v_n\}$ be two sequences of real numbers and suppose that
i) $v_n > 0$ for all $n$ and $\displaystyle \sum_{n=1}^{\infty} v_n=\infty,$
ii) $\displaystyle \lim_{n\to\infty} \frac{u_n}{v_n}=\ell.$
Show that $\displaystyle \lim_{n\to\infty} \frac{u_1 + \cdots + u_n}{v_1 + \cdots + v_n}=\ell.$

Solution. For $n \ge 1,$ let $\displaystyle a_n:=\sum_{k=1}^nu_k, \ b_n:=\sum_{k=1}^n v_k$ and apply the above problem. $\Box$

Example 2. Let $\{x_n\}$ be a sequence of real numbers and suppose that $\displaystyle \lim_{n\to\infty}x_n=x.$ Show that $\displaystyle \lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n x_k=x$ and, if $x_n > 0$ for all $n,$ $\displaystyle \lim_{n\to\infty} \sqrt[n]{x_1 x_2 \cdots x_n}=x.$

Solution. Both parts follow trivially from Example 1: for the first part, consider the sequences $u_n:=x_n, \ v_n:=1$ and for the second part, consider the sequences $u_n:=\ln x_n, \ v_n:=1. \ \Box$

Example 3 (see also this post!) . Show that $\displaystyle \lim_{n\to\infty} \frac{\sqrt[n]{n!}}{n}=\frac{1}{e}.$

Solution. Let $\displaystyle x_n:=\left(1+\frac{1}{n}\right)^n, \ n \ge 1,$ and apply the second part of Example 2. $\Box$

Example 4. Show that $\displaystyle \lim_{n\to\infty} \frac{1}{n^n} \sum_{k=1}^n k^k = 1.$

Solution. Let $\displaystyle a_n:=\sum_{k=1}^n k^k, \ b_n=n^n, \ n \ge 1.$ Then $\displaystyle \lim_{n\to\infty} \frac{a_n-a_{n-1}}{b_n-b_{n-1}}=\lim_{n\to\infty} \frac{n^n}{n^n-(n-1)^{n-1}}=\lim_{n\to\infty} \frac{1}{1-\left(1-\frac{1}{n}\right)^n(n-1)^{-1}}=1,$
because $\displaystyle \lim_{n\to\infty} \left(1-\frac{1}{n}\right)^n=e^{-1}$ and $\displaystyle \lim_{n\to\infty} (n-1)^{-1}=0.$ The result now follows from the Stolz–Cesàro lemma. $\Box$

Exercise 1. Prove the Stolz–Cesàro lemma for $\ell = \pm \infty.$

Exercise 2. Let $\{x_n\}$ be a sequence of real numbers and suppose that $\displaystyle \lim_{n\to\infty}nx_n=\ell.$ Show that $\displaystyle \lim_{n\to\infty} \frac{1}{\ln n}\sum_{k=1}^nx_k=\ell.$

## Limit of integrals (8)

Problem 1. Consider the sequence $\displaystyle I_n:=\frac{1}{n!}\int_n^{\infty}x^ne^{-x}dx.$ Show that $\displaystyle \lim_{n\to\infty}I_n=\frac{1}{2}.$

Solution. My solution is basically the same as how I proved Stirling’s formula; so you should consider this problem as a nice exercise of that post.
We begin with the substitution $x=n(1+t)$ to get $\displaystyle I_n=\frac{n^{n+1}e^{-n}}{n!}\int_0^{\infty}e^{nf(t)}dt,$ where $f(t):=\ln(1+t)-t.$ Now choose any real number $a$ such that $\displaystyle \frac{1}{3} < a < \frac{1}{2}.$ Then
$\displaystyle I_n=\frac{n^{n+\frac{1}{2}}e^{-n}}{n!}\left(\sqrt{n}\int_0^{\frac{1}{n^a}}e^{nf(t)}dt + \sqrt{n}\int_{\frac{1}{n^a}}^{\infty}e^{nf(t)}dt \right). \ \ \ \ \ \ \ \ \ \ \ \ (\star)$
Now, by Stirling’s formula, we have $\displaystyle \lim_{n\to\infty} \frac{n^{n+\frac{1}{2}}e^{-n}}{n!}=\frac{1}{\sqrt{2\pi}}.$ Also, by ii), iii) in the solution of Problem 2, $\displaystyle \lim_{n\to\infty} \sqrt{n}\int_{\frac{1}{n^a}}^{\infty}e^{nf(t)}dt=0$ and $\displaystyle \lim_{n\to\infty} \sqrt{n}\int_0^{\frac{1}{n^a}}e^{nf(t)}dt =\frac{\sqrt{2\pi}}{2}.$ The result now follows from $(\star). \ \Box$

Example. Show that $\displaystyle \lim_{n\to\infty} e^{-n}\sum_{k=0}^n \frac{n^k}{k!}=\frac{1}{2}.$

Solution. We show that $\displaystyle e^{-n}\sum_{k=0}^n \frac{n^k}{k!}=\frac{1}{n!}\int_n^{\infty}x^ne^{-x}dx$ and so the limit is $\displaystyle \frac{1}{2},$ by Problem 1. To show that, consider the sequence $\displaystyle c_m:=\frac{1}{m!}\int_n^{\infty}x^me^{-x}dx.$ Then $c_0=e^{-n}$ and integration by parts with $x^m=u, \ s^{-x}dx=dv$ gives $\displaystyle c_m=\frac{n^m}{m!}e^{-n}+c_{m-1}.$ Thus $\displaystyle c_m=e^{-n}\sum_{k=0}^m \frac{n^k}{k!}$ and the result follows. $\Box$

Remark. If $m \ge 0$ is an integer independent of $n,$ then $\displaystyle \lim_{n\to\infty} e^{-n}\sum_{k=0}^m \frac{n^k}{k!}=0$ because $\displaystyle \sum_{k=0}^m \frac{n^k}{k!}$ is a polynomial in $n$ and we know that if $p(x)$ is a polynomial, then $\displaystyle \lim_{x\to\infty} \frac{p(x)}{e^x}=0.$ This fact also follows, trivially, from $\displaystyle e^{-n}\sum_{k=0}^m \frac{n^k}{k!}=\frac{1}{m!}\int_n^{\infty} x^me^{-x}dx,$ which we proved in the above example.

Problem 2. Show that the sequence in Problem 1 is decreasing.

Solution. Integration by parts with $x^{n+1}=u$ and $e^{-x}dx=dv$ gives $\displaystyle (n+1)!I_{n+1}=\int_{n+1}^{\infty}x^{n+1}e^{-x}dx=(n+1)^{n+1}e^{-(n+1)}+(n+1)\int_{n+1}^{\infty}x^ne^{-x}dx$
and thus
$\displaystyle n!I_{n+1}=(n+1)^ne^{-(n+1)}+\int_{n+1}^{\infty}x^ne^{-x}dx. \ \ \ \ \ \ \ \ \ \ \ (\star \star)$
Now see that the function $g(x):=x^ne^{-x}$ is decreasing on the interval $[n,n+1]$ and so $\displaystyle (n+1)^ne^{-(n+1)}=g(n+1) < \int_n^{n+1}g(x) \ dx=\int_n^{n+1}x^ne^{-x}dx.$ Thus, by $(\star \star),$ $\displaystyle n!I_{n+1} < \int_n^{n+1}x^ne^{-x}dx + \int_{n+1}^{\infty}x^ne^{-x}dx=\int_n^{\infty}x^ne^{-x}dx=n!I_n. \ \Box$

Exercise 1. Show that the sequence $\displaystyle J_n:= \frac{1}{n!}\int_0^n x^ne^{-x}dx$ is increasing and $\displaystyle \lim_{n\to\infty}J_n=\frac{1}{2}.$

Exercise 2. Show that $\displaystyle e^n < 2\sum_{k=0}^n \frac{n^k}{k!}$ for all $n.$

Exercise 3. Let $I_n$ and $J_n$ be the sequences in Problem 1 and Exercise 1, respectively. True or false: the sequences $n!I_n$ and $n!J_n$ are both increasing?

## The sequence n!^(1/n)/n

Consider the sequence $\displaystyle u_n:=\frac{\sqrt[n]{n!}}{n}, \ n \ge 1.$ It is quite easy to show that $\displaystyle \lim_{n\to\infty} u_n=e^{-1}$ (Problem 2, i)) A more interesting fact is that $\{u_n\}$ is decreasing (Problem 2, ii)). But before proving that, we need to prove something that in this post I gave as exercise.

Problem 1. Show that the sequence $\displaystyle b_n:=\left(1+\frac{1}{n} \right)^{n+1}, \ n \ge 1,$ is decreasing and so $\displaystyle b_n > e$ for all $n \ge 1.$

Solution. We have
$\displaystyle (b_{n+1})^{\frac{1}{n+1}}=\left(1+\frac{1}{n+1} \right)^{\frac{n+2}{n+1}}=\left(1+\frac{1}{n+1} \right)\left(1+\frac{1}{n+1} \right)^{\frac{1}{n+1}}$
and so, by the Bernoulli’s inequality, $\displaystyle (b_{n+1})^{\frac{1}{n+1}} < \left(1+\frac{1}{n+1} \right)\left(1+\frac{1}{(n+1)^2} \right) < \left(1+\frac{1}{n+1} \right)\left(1+\frac{1}{n^2+2n} \right) =1+\frac{1}{n}.$
Thus $\displaystyle b_{n+1} < \left(1+\frac{1}{n}\right)^{n+1}=b_n. \ \Box$

Problem 2. Consider the sequence $\displaystyle u_n:=\frac{\sqrt[n]{n!}}{n}, \ n \ge 1.$ Show that
i) $\displaystyle \lim_{n\to\infty} u_n=e^{-1},$
ii) $\displaystyle e^{n-1} \ge \frac{n^n}{n!}$ for $n \ge 1.$ The equality holds for $n=1$ only,
iii) $\displaystyle \{u_n\}$ is decreasing.

Solution. i) Since $\displaystyle \ln u_n=\frac{1}{n}\ln n! - \ln n=\frac{1}{n}\sum_{k=1}^n \ln \left(\frac{k}{n}\right)$ and $\ln x$ is a continuous function, we have
$\displaystyle \ln \left(\lim_{n\to\infty} u_n \right)=\lim_{n\to\infty} \ln u_n = \lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n \ln \left(\frac{k}{n}\right)=\int_0^1 \ln x \ dx=-1.$

ii) For $n=1$ we have an equality.
First Solution. We use induction over $n \ge 2.$ The inequality holds for $n=2$ because $e > 2.$ Now suppose that it holds for $n.$ Then, using Example 2 in this post, we have
$\displaystyle e^n =e \cdot e^{n-1} > e \cdot \frac{n^n}{n!} > \left(1+\frac{1}{n}\right)^n \frac{n^n}{n!}=\frac{(n+1)^{n+1}}{(n+1)!}.$
Second Solution. If $n > 1,$ then $\displaystyle e^n =\sum_{k=0}^{\infty} \frac{n^k}{k!} > \sum_{k=n-1}^{\infty} \frac{n^k}{k!}=\frac{n^n}{n!}s_n,$ where
$\displaystyle s_n=2 + \sum_{k=1}^{\infty} \frac{n^k}{(n+1)(n+2) \cdots (n+k)}=2+\sum_{k=1}^{\infty} \frac{1}{(1+\frac{1}{n})(1+\frac{2}{n}) \cdots (1+\frac{k}{n})} > 2 + \sum_{k=1}^{\infty} \frac{1}{(k+1)!}=e.$

iii) Using ii) and Problem 1, we have
$\displaystyle \left(\frac{u_n}{u_{n+1}}\right)^{n(n+1)}=\frac{n!}{n^n}\left(1+\frac{1}{n}\right)^{n^2} > e^{1-n} \left(1+\frac{1}{n}\right)^{n^2-1} = e^{1-n} \left[\left(1+\frac{1}{n}\right)^{n+1}\right]^{n-1} > e^{1-n}e^{n-1}=1. \ \Box$

Remark. Unlike the inequality $\displaystyle e^{n-1} > \frac{n^n}{n!},$ the inequality  $\displaystyle e^n > \frac{n^n}{n!}$ is trivial because $\displaystyle e^n=\sum_{k=0}^{\infty} \frac{n^k}{k!} > \frac{n^n}{n!}.$

Exercise. Show that $\displaystyle \lim_{n\to\infty} \sqrt[n]{\sum_{k=0}^n \frac{n^k}{k!}}=e.$

## Irrationality of pi

We have already proved that $e$ is irrational (see the example in this post!)
The fact that $\pi$ is irrational, i.e. there are no integers $a,b$ such that $\displaystyle \pi=\frac{a}{b},$ was discovered a couple of hundreds of years ago and there are several proofs of this result; I like Ivan Niven’s proof better and that’s what I’m going to explain here.

Throughout this post, $f^{(k)}(x)$ is the $k$-th derivative of a function $f(x).$

Problem 1. Let $P(x)$ be a polynomial of degree $2n.$ Show that
$\displaystyle \int_0^{\pi} P(x) \sin x \ dx = \sum_{k=0}^n (-1)^k(P^{(2k)}(0)+P^{(2k)}(\pi)).$

Solution. The proof is by induction over $n.$ If $n=0,$ then $P(x)=p$ is a constant and so $\displaystyle \int_0^{\pi} P(x) \sin x \ dx=p\int_0^{\pi} \sin x \ dx = 2p=P(0)+P(\pi).$
Now suppose the equality in the problem holds for polynomials of degree $2n$ and let $P(x)$ be a polynomial of degree $2n+2.$ Then using integration by parts with $P(x)=u$ and $\sin x \ dx = dv$ gives $\displaystyle \int_0^{\pi} P(x) \sin x \ dx = P(0)+P(\pi)+\int_0^{\pi}P'(x)\cos x \ dx. \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$
Now, in $(1)$, we use integration by parts again, this time with $P(x)=u$ and $\cos x \ dx = dv$ to get $\displaystyle \int_0^{\pi} P(x) \sin x \ dx = P(0)+P(\pi)-(P'(0)+P'(\pi)) + \int_0^{\pi} P''(x) \sin x \ dx. \ \ \ \ \ \ \ \ \ \ \ \ (2)$
But since $P''(x)$ is a polynomial of degree $2n,$ we can use our induction hypothesis to write $\displaystyle \int_0^{\pi} P''(x) \sin x \ dx= \sum_{k=0}^n (-1)^k(P^{(2k+2)}(0)+P^{(2k+2)}(\pi))$ and so $(2)$ becomes
$\displaystyle \int_0^{\pi} P(x) \sin x \ dx = P(0)+P(\pi)-(P'(0)+P'(\pi)) + \sum_{k=0}^n (-1)^k(P^{(2k+2)}(0)+P^{(2k+2)}(\pi))= \sum_{k=0}^{n+1} (-1)^k(P^{(2k)}(0)+P^{(2k)}(\pi)). \ \Box$

Problem 2. Let $n,a,b$ be integers with $n \ge 1$ and $b \ne 0.$ Let $\displaystyle c:=\frac{a}{b}$ and $\displaystyle P(x):=\frac{x^n(a-bx)^n}{n!}.$ Show that $P^{(k)}(0)$ and $P^{(k)}(c)$ are integers for all $k \ge 0.$

Solution. Since $P(x)$ is a polynomial of degree $2n,$ we have $P^{(k)}(x)=0$ for all $x$ and $k > 2n$ and so there is nothing to prove for $k > 2n.$
Now, since $P(x)$ is a polynomial of degree $2n,$ we have
$\displaystyle P(x)=\sum_{k=0}^{2n} \frac{P^{(k)}(0)}{k!}x^k. \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$
On the other hand, using the binomial theorem, it is clear that
$\displaystyle P(x)=\frac{x^n(a-bx)^n}{n!}=\sum_{k=n}^{2n}\frac{p_k}{n!}x^k, \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$
where $p_k$ are some integers.
So $(1)$ and $(2)$ together give $P^{(k)}(0)=0$ for $0 \le k \le n-1$ and $\displaystyle P^{(k)}(0)=\frac{k!}{n!}p_k$ for $n \le k \le 2n.$ Thus, since $\displaystyle \frac{k!}{n!}$ is an integer for $k \ge n,$ we have proved that $P^{(k)}(0)$ is an integer for all $k \ge 0.$
Finally, since $P(c-x)=P(x),$ we have $(-1)^kP^{(k)}(c-x)=P^{(k)}(x)$ for all $k \ge 0$ and thus $(-1)^kP^{(k)}(0)=P^{(k)}(c).$ So $P^{(k)}(c)$ is an integer because we have already proved that $P^{(k)}(0)$ is an integer. $\Box$

Problem 3. Show that $\pi$ is irrational.

Solution (Ivan Niven). Suppose, to the contrary, that $\pi$ is rational and put $\displaystyle \pi=\frac{a}{b},$ where $a,b > 0$ are integers. Let $n \ge 1$ be an integer and put
$\displaystyle P(x):=\frac{x^n(a-bx)^n}{n!}.$
Then by Problem 1 and 2, $\displaystyle \int_0^{\pi} P(x) \sin x \ dx$ must be an integer. But we are now going to prove that if $n$ is large enough, then $\displaystyle \int_0^{\pi} P(x) \sin x \ dx$ is not an integer and this contradiction proves that our assumption that $\pi$ is rational is false.
First note that, on the interval $[0,\pi],$ we have
$\displaystyle 0 \le P(x)\sin x \le P(x) =\frac{b^nx^n(\pi - x)^n}{n!} \le \frac{b^n \pi^{2n}}{n!}=\frac{(b \pi^2)^n}{n!}$
and thus
$\displaystyle 0 < \int_0^{\pi} P(x)\sin x \ dx \le \frac{(b \pi^2)^n}{n!} \pi. \ \ \ \ \ \ \ \ \ \ (*)$
(Note that $\displaystyle \int_0^{\pi} P(x)\sin x \ dx \ne 0$ because $P(x) \sin x$ is not identically zero on $[0, \pi]$).
But for every real number $r,$ we have $\displaystyle \lim_{n\to\infty} \frac{r^n}{n!}=0$ because the series $\displaystyle \sum_{n=0}^{\infty} \frac{r^n}{n!}$ is convergent (to $e^r$). So $\displaystyle \lim_{n\to\infty} \frac{(b \pi^2)^n}{n!} \pi=0.$ Hence, if $n$ is large enough, we will have $\displaystyle \frac{(b \pi^2)^n}{n!} \pi < 1$ and thus, by $(*),$ $\displaystyle 0 < \int_0^{\pi} P(x)\sin x \ dx < 1.$ Therefore $\displaystyle \int_0^{\pi} P(x)\sin x \ dx$ is not an integer if $n$ is large enough. $\Box$

Remark. Another way to say that a real number $t$ is irrational is to say that $t$ is not the root of a polynomial of degree one with integer coefficients. If $t$ is not a root of any polynomial with integer coefficients, then $t$ is called transcendental. Clearly every transcendental number is irrational. It is known that both $e$ and $\pi$ are in fact transcendental.

Exercise 1. Let $P(x)$ be a polynomial of degree $2n+1.$ Show that
$\displaystyle \int_0^{\pi} P(x) \sin x \ dx = \sum_{k=0}^n (-1)^k(P^{(2k)}(0)+P^{(2k)}(\pi)).$

Exercise 2. Show that the inequality $(*)$ in the solution of Problem 3 can be improved by proving that $\displaystyle \int_0^{\pi} P(x)\sin x \ dx \le \frac{(b \pi^2)^n}{4^n n!} \pi.$

## Limit of (1^n + 2^n + … + n^n)/n^n

We have all seen this easy exercise: show that $\displaystyle \lim_{n\to\infty} \frac{1^{\alpha} + 2^{\alpha} + \cdots + n^{\alpha}}{n^{\alpha + 1}}=\frac{1}{\alpha +1},$ where $\alpha > -1$ is a real constant, i.e. $\alpha$ is independent of $n.$ The proof is quite simple: we have $\displaystyle \frac{1^{\alpha} + 2^{\alpha} + \cdots + n^{\alpha}}{n^{\alpha + 1}}= \frac{1}{n}\sum_{k=1}^n \left(\frac{k}{n}\right)^{\alpha},$ which is a Riemann sum, and so the limit is $\displaystyle \int_0^1 x^{\alpha} dx = \frac{1}{\alpha +1}.$
But things get interesting when $\alpha$ depends on $n.$
For example, what is $\displaystyle \lim_{n\to\infty} \frac{1^n + 2^n + \cdots + n^n}{n^{n + 1}} ?$ Well, the answer is $0$ and that’s a trivial consequence of a much nicer result that we are going to prove in Problem 2. But first, we need to prove something, which is nice by itself.

Problem 1. If $r$ is a real constant, show that $\displaystyle \lim_{n\to\infty} \sum_{k=1}^{n-1} \frac{k^re^{-k}}{n-k}=0.$

Solution. Choose an integer $m \ge 0$ such that $r \le m.$ Then $\displaystyle e^k > \frac{k^{m+1}}{(m+1)!} \ge \frac{k^{r+1}}{(m+1)!}$ and so
$\displaystyle 0 < \sum_{k=1}^{n-1} \frac{k^re^{-k}}{n-k} < (m+1)! \sum_{k=1}^{n-1} \frac{1}{k(n-k)}=\frac{2(m+1)!}{n}\sum_{k=1}^{n-1}\frac{1}{k} < \frac{2(m+1)!}{n}\sum_{k=1}^n\frac{1}{k}=\frac{2(m+1)!}{n}\left(\sum_{k=1}^n\frac{1}{k}-\ln n \right) + \frac{2(m+1)! \ln n}{n}.$
Now the result follows from the facts that $\displaystyle \lim_{n\to\infty} \frac{1}{n}\left(\sum_{k=1}^n\frac{1}{k}-\ln n \right)=\lim_{n\to\infty}\frac{1}{n}\gamma=0,$ where $\gamma$ is the Euler’s constant, and $\displaystyle \lim_{n\to\infty} \frac{\ln n}{n}=0. \ \Box$

Problem 2. Let $\displaystyle a_n:= \frac{1^n + 2^n + \cdots + n^n}{n^n}.$ Show that $\displaystyle \lim_{n\to\infty} a_n=\frac{e}{e-1}.$

Solution. By part i) and the Example in this problem, we have
$\displaystyle e^{\frac{x}{1+x}} \le 1+ x \le e^x, \ \ \ \ \ \ \ \ \ \ \ \ \ \ (*)$
for all real numbers $x.$ Thus
$\displaystyle a_n =\sum_{k=1}^n \left(\frac{k}{n}\right)^n=\sum_{k=0}^{n-1} \left(1-\frac{k}{n}\right)^n\le \sum_{k=0}^{n-1}e^{-k} \le \sum_{k=0}^{\infty}e^{-k}=\frac{e}{e-1}.$
On the other hand, again by $(*),$ we have
$\displaystyle a_n=\sum_{k=0}^{n-1} \left(1-\frac{k}{n}\right)^n \ge \sum_{k=0}^{n-1} e^{\frac{-kn}{n-k}}=\sum_{k=0}^{n-1}e^{-k} - \sum_{k=0}^{n-1} \left(1-e^{\frac{-k^2}{n-k}} \right)e^{-k} \ge \sum_{k=0}^{n-1}e^{-k} - \sum_{k=0}^{n-1} \frac{k^2e^{-k}}{n-k}.$
So we have proved that $\displaystyle \sum_{k=0}^{n-1}e^{-k} - \sum_{k=0}^{n-1} \frac{k^2e^{-k}}{n-k} \le a_n \le \frac{e}{e-1}.$ Hence, in order to complete the solution, we only need to show that $\displaystyle \lim_{n\to\infty} \left(\sum_{k=0}^{n-1}e^{-k} - \sum_{k=0}^{n-1} \frac{k^2e^{-k}}{n-k}\right)=\frac{e}{e-1}$
and that’s easy to do since $\displaystyle \lim_{n\to\infty} \sum_{k=0}^{n-1}e^{-k}=\sum_{k=0}^{\infty}e^{-k}=\frac{e}{e-1}$ and, by Problem 1, $\displaystyle \lim_{n\to\infty} \sum_{k=0}^{n-1} \frac{k^2e^{-k}}{n-k}=0. \ \Box$

Remark. There’s a shorter proof of Problem 2 that uses Tannery’s theorem but I don’t discuss that theorem in this blog (I like my solution better!)

Exercise 1. If $\alpha \le -1$ is a real constant, what is $\displaystyle \lim_{n\to\infty} \frac{1^{\alpha} + 2^{\alpha} + \cdots + n^{\alpha}}{n^{\alpha + 1}} ?$

Exercise 2. Show that $\displaystyle \sum_{k=0}^n \binom{n+1}{k}\frac{B_k}{n^k}=\frac{1}{e-1},$ where $B_k$ are Bernoulli numbers.
Hint. Use Problem 2 and the problem in this post.

Exercise 3. Prove that $\displaystyle \lim_{n\to\infty} \frac{1^n + 2^n + \cdots + n^n}{n^{n + 1}}=0$ without using the result given in Problem 2.
Hint. You’ll only need the easy half of the solution of Problem 2.

## Convergence of series (2)

Problem 1. Show that if $\displaystyle \sum_{n=1}^{\infty} a_n, \ 0\le a_n < 1,$ converges, then $\displaystyle \sum_{n=1}^{\infty} \frac{a_n}{1-a_n}$ converges too.

Solution. Since $\displaystyle \lim_{n\to\infty} a_n=0,$ there exists an integer $N > 0$ such that $\displaystyle a_n \le \frac{1}{2}$ for all $n \ge N.$  Then $\displaystyle 1-a_n \ge \frac{1}{2}$ for $n \ge N$ and so $\displaystyle 0 \le \frac{a_n}{1-a_n} \le 2a_n$ for $n \ge N.$ The result now follows from the comparison test. $\Box$

Problem 2. Show that
i) if $\displaystyle \sum_{n=1}^{\infty} a_n, \ a_n \ge 0,$ converges, then $\displaystyle \sum_{n=1}^{\infty} \ln(1+a_n)$ converges too.
ii) if $\displaystyle \sum_{n=1}^{\infty} a_n, \ 0 \le a_n < 1,$ converges, then $\displaystyle \sum_{n=1}^{\infty} \ln(1-a_n)$ converges too.

Solution. By the Example in this post, we have $\displaystyle \frac{x}{x+1} \le \ln(x+1) \le x$ for all real numbers $x > -1.$ So $\displaystyle 0 \le \ln(1+a_n) \le a_n$ for $a_n \ge 0$ and $\displaystyle 0 \le -\ln(1-a_n) \le \frac{a_n}{1-a_n}$ for $0 \le a_n < 1.$ The result now follows from the comparison test and Problem 1. $\Box$

Exercise 1. Show that if $\displaystyle \sum_{n=1}^{\infty} a_n, \ a_n \ge 0,$ diverges, then $\displaystyle \sum_{n=1}^{\infty} \frac{a_n}{a_n+1}$ diverges too.
Hint. Suppose that $\displaystyle \sum_{n=1}^{\infty} \frac{a_n}{a_n+1}$ converges and then use Problem 1.

Exercise 2. Show that $\displaystyle \sum_{n=1}^{\infty} \ln \left(\cos \left(\frac{1}{n}\right) \right)$ is convergent.

## Integral of ln(x^2 + a)/(x^2 + 1)

We have already seen one example of evaluating definite integrals using the Leibniz integral rule. In this post, I give another one.

Problem. Show that $\displaystyle \int_0^{\infty} \frac{\ln(x^2+a)}{x^2+1} \ dx = \pi \ln(\sqrt{a}+1)$ for all real constants $a \ge 0.$

Solution. Let $\displaystyle f(a):=\int_0^{\infty} \frac{\ln(x^2+a)}{x^2+1} \ dx, \ a \ge 0.$
First see that the equality in the problem holds for $a=0$ because if, in $\displaystyle f(0)=2\int_0^{\infty} \frac{\ln x}{x^2+1} \ dx,$ we put $\displaystyle x = \frac{1}{t},$ then we will quickly get $\displaystyle f(0)=0.$
So, from now on, we will assume that $a > 0.$ Now we use the definition of derivative to find $f'(a),$ the derivative of $f$ with respect to $a.$ We have $\displaystyle f'(a)=\lim_{h\to0} \frac{1}{h} \int_0^{\infty} \frac{1}{x^2+1} \ln \left(1+\frac{h}{x^2+a} \right) dx. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$
Now we have $\displaystyle \frac{t}{t+1} \le \ln(1+t) \le t$ for all real numbers $t > -1$ (see the Example in this post), and thus
$\displaystyle \frac{h}{x^2+a+h} \le \ln \left(1+\frac{h}{x^2+a} \right) \le \frac{h}{x^2+a}. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$
It now follows from $(1)$ and $(2)$ that
$\displaystyle f'(a)=\int_0^{\infty} \frac{dx}{(x^2+a)(x^2+1)}=\frac{\pi}{2(a+\sqrt{a})}$
and therefore
$\displaystyle f(a)=\frac{\pi}{2} \int \frac{da}{a+\sqrt{a}}=\pi \ln(\sqrt{a}+1) + C, \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$
for some constant $C.$
But what is $C?$ Well, if in $(3)$ we, for example, put $a = 1,$ then we’ll get
$C=f(1)-\pi \ln 2. \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)$
We also have $\displaystyle f(1)=\int_0^{\infty} \frac{\ln(x^2+1)}{x^2+1} \ dx=\int_0^{\frac{\pi}{2}} \ln(\tan^2 \theta + 1) \ d \theta = -2 \int_0^{\frac{\pi}{2}} \ln(\cos \theta) \ d \theta=-2 \int_0^{\frac{\pi}{2}} \ln(\sin \theta) \ d \theta$
and so $f(1)=\pi \ln 2,$ by this problem. Thus, by $(4), \ C=0$ and we are done by $(3). \ \Box$

Example. Show that $\displaystyle g(c):=\int_0^{\pi} \ln(c+\cos x) \ dx = \pi \ln \left(\frac{c+\sqrt{c^2-1}}{2}\right)$ for all real constants $c \ge 1.$

Solution. We first prove the equality for $c=1.$ We have $\displaystyle g(1)=\int_0^{\pi} \ln(1+\cos x) \ dx = \int_0^{\pi} \left(\ln2 + 2\ln \cos \left(\frac{x}{2}\right) \right) \ dx=\pi \ln 2 + 4 \int_0^{\frac{\pi}{2}} \ln \cos x \ dx \\ =\pi \ln 2 + 4 \int_0^{\frac{\pi}{2}} \ln \sin x \ dx.$
Thus $g(1)=-\pi \ln 2,$ by this problem, and we’re done in this case.
Now suppose that $c > 1$ and let $\displaystyle \tan \left(\frac{x}{2}\right)=y.$ Then $\displaystyle \cos x = \frac{1-y^2}{1+y^2}$ and $\displaystyle dx = \frac{2dy}{1+y^2}.$ Hence, by the above problem, we have
$\displaystyle g(c)=2 \int_0^{\infty} \frac{\ln(c-1) + \ln \left(y^2 + \frac{c+1}{c-1} \right) - \ln(y^2+1)}{y^2+1} \ dy=\pi \ln(c-1) + 2\pi \ln \left(\sqrt{\frac{c+1}{c-1}}+1\right)-2\pi \ln 2=\pi \ln \left(\frac{c+\sqrt{c^2-1}}{2}\right). \ \Box$

Exercise 1. Show that $\displaystyle \int_0^{\infty} \frac{dx}{(x^2+a)(x^2+1)}=\frac{\pi}{2(a+\sqrt{a})}$ for all real constants $a > 0.$

Exercise 2. For a real constant $c \ge 1,$ show that  $\displaystyle \int_0^{\frac{\pi}{2}} \ln(c^2-\sin^2x) \ dx=\int_0^{\pi} \ln(c+\cos x) \ dx.$

Exercise 3. Evaluate $\displaystyle \int_0^{\infty} \frac{\ln(x^2+a)}{x^2+b} \ dx,$ where $a \ge 0$ and $b > 0$ are real constants.

## Bernoulli’s inequality

If $x \ge 0$ is a real number and $n \ge 1$ is an integer, then it is clear, from the binomial theorem, that $\displaystyle (1+x)^n \ge 1+\binom{n}{1}x=1+nx.$ This is the trivial case of the Bernoulli’s inequality.
Let’s extend that result. Let $r, x$ be real numbers with $x > -1$ (we need the condition $x > -1$ to make sure that $(1+x)^r$ is a real number). When do we have $(1+x)^r \ge 1+rx$ or $(1+x)^r \le 1+rx ?$ For $x=0$ or $r=0,1,$ we have $(1+x)^r=1+rx$ and so we will assume that $x \ne 0$ and $r \ne 0,1.$

Problem (Bernoulli’s inequality). Let $r,x$ be real numbers with $r \ne 0,1$ and $0 \ne x > -1.$ Show that
i) if $r > 1$ or $r < 0,$ then $(1+x)^r > 1+rx.$
ii) if $0 < r < 1,$ then $(1+x)^r < 1+rx.$

Solution. Given a real number $r,$ let $f(x):=(1+x)^r.$ By Taylor’s theorem, for every nonzero $x > -1,$ there exists $c$ between $0$ and $x$ such that $\displaystyle f(x)=f(0)+f'(0)x+\frac{f''(c)}{2}x^2.$
Thus, since $f(0)=1, \ f'(0)=r$ and $f''(c)=r(r-1)(1+c)^{r-2},$ we have
$\displaystyle (1+x)^r=1+rx+\frac{r(r-1)}{2}(1+c)^{r-2}x^2$
and hence
$\displaystyle (1+x)^r -(1+rx)=\frac{r(r-1)}{2}(1+c)^{r-2}x^2. \ \ \ \ \ \ \ \ \ \ \ \ \ (*)$
Now $c > -1$ because $x > -1$ and $c$ is between $0$ and $x.$ Thus $(1+c)^{r-2} > 0$ and so, by $(*),$ if $r(r-1) > 0,$ i.e. if $r > 1$ or $r < 0,$ then $(1+x)^r-(1+rx) > 0$ and if $r(r-1) < 0,$ i.e. if $0 < r < 1,$ then $(1+x)^r-(1+rx) < 0. \ \Box$

Example 1. Show that $\displaystyle \lim_{r\to0+} \sum_{n=1}^{\infty} \frac{n^r}{n!}=e-1.$

Solution. (I have also given this solution here) By part ii) of the above problem, $1 \le (n+1)^r \le 1+nr$ for $n \ge 0$ and real numbers $0 < r < 1.$ So for $0 < r < 1,$
$\displaystyle \sum_{n=0}^{\infty} \frac{1}{(n+1)!} \le \sum_{n=0}^{\infty}\frac{(n+1)^r}{(n+1)!} \le \sum_{n=0}^{\infty} \frac{1}{(n+1)!}+r \sum_{n=0}^{\infty} \frac{n}{(n+1)!}.$
But $\displaystyle \lim_{r\to0} r \sum_{n=0}^{\infty} \frac{n}{(n+1)!}=0,$ because $\displaystyle \sum_{n=0}^{\infty} \frac{n}{(n+1)!}$ is convergent, and so, by the squeeze theorem, $\displaystyle \lim_{r\to0+} \sum_{n=1}^{\infty} \frac{n^r}{n!}=\sum_{n=0}^{\infty} \frac{1}{(n+1)!}=e-1. \ \Box$

Example 2. Show that the sequence $\displaystyle a_n:=\left(1+\frac{1}{n}\right)^n, \ \ n\ge 1,$ is increasing and so $a_n < e$ for $n \ge 1.$

Solution. We have $\displaystyle \frac{a_{n+1}}{a_n}=\frac{\left(1+\frac{1}{n+1}\right)^{n+1}}{\left(1+\frac{1}{n}\right)^n}=\left(1+\frac{1}{n}\right)\left(\frac{1+\frac{1}{n+1}}{1+\frac{1}{n}}\right)^{n+1} = \left(1+\frac{1}{n}\right) \left(\frac{n^2+2n}{(n+1)^2} \right)^{n+1}=\left(1+\frac{1}{n}\right) \left(1-\frac{1}{(n+1)^2} \right)^{n+1}.$
But by part i) of the above problem, $\displaystyle \left(1-\frac{1}{(n+1)^2} \right)^{n+1} > 1-\frac{1}{n+1}$ and thus $\displaystyle \frac{a_{n+1}}{a_n} > \left(1+\frac{1}{n}\right)\left(1-\frac{1}{n+1}\right)=1. \ \Box$

Example 3 (AM-GM inequality). Show that $\displaystyle \frac{x_1+x_2 + \cdots + x_n}{n} \ge \sqrt[n]{x_1x_2 \cdots x_n}$ for any integer $n \ge 1$ and positive real numbers $x_1, x_2, \cdots , x_n.$

Solution. First note that, by part i) of the above problem, $x^r \ge 1+r(x-1)$ for $x > 0$ and $r \ge 1$ (just change $x$ to $x-1$).
Now, back to the solution, there’s nothing to prove for $n=1$ and so we’ll assume that $n\ge 2.$ For integers $1 \le k \le n,$ let $\displaystyle \sigma_k:=\frac{x_1+x_2 + \cdots + x_k}{k}.$ We want to show that $\sigma_n^n \ge x_1x_2 \cdots x_n.$
If $k \ge 2,$ then, as we mentioned at the beginning of the solution, $\displaystyle \left(\frac{\sigma_k}{\sigma_{k-1}}\right)^k \ge 1 + k \left(\frac{\sigma_k}{\sigma_{k-1}}-1 \right)=\frac{k\sigma_k-(k-1)\sigma_{k-1}}{\sigma_{k-1}}=\frac{x_k}{\sigma_{k-1}}$
and so $\displaystyle \sigma_k^k \ge x_k \sigma_{k-1}^{k-1}.$ Thus we have
$\displaystyle \sigma_n^n \ge x_n \sigma_{n-1}^{n-1} \ge x_nx_{n-1}\sigma_{n-2}^{n-2} \ge \cdots \ge x_nx_{n-1} \cdots x_2 \sigma_1=x_nx_2 \cdots x_1. \ \Box$

Exercise.  Show that the sequence $\displaystyle b_n:=\left(1+\frac{1}{n}\right)^{n+1}, \ \ n\ge 1,$ is decreasing and so $\displaystyle \left(1+\frac{1}{n}\right)^n < e < \left(1+\frac{1}{n}\right)^{n+1}$ for $n \ge 1.$
Hint. You can always see this post for a solution.

## Stirling’s formula (1)

Stirling’s formula (Problem 2) gives a useful asymptotic approximation for $n!.$ Using this formula, we can easily evaluate many limits involving factorials.

Problem 1. Let $a>0$ and $0 < b < 1$ be real numbers and consider the function $\displaystyle f(t):=\ln(t+1)-t, \ t>-1.$
Show that
i) $\displaystyle f(t) < \frac{-b^{2a}}{2}$ for $t \in (-1, -b^a]$ and $\displaystyle f(t) < \frac{-b^a}{6}t$ for $\displaystyle t \ge b^{a}.$
ii) $\displaystyle - \frac{t^2}{2} - \frac{b^{3a}}{3(1-b^a)^3} < f(t) < - \frac{t^2}{2} + \frac{b^{3a}}{3(1-b^a)^3}$ for $\displaystyle t \in [-b^a, b^a].$

Solution. By Taylor’s theorem, for every $t > -1$ there exists some $c$ between $0$ and $t$ such that $\displaystyle f(t)=f(0)+f'(0)t + \frac{f''(0)}{2}t^2+\frac{f'''(c)}{6}t^3.$ Thus
$\displaystyle f(t)=-\frac{t^2}{2}+\frac{t^3}{3(c+1)^3}. \ \ \ \ \ \ \ \ \ \ \ \ \ (*)$
Now back to the problem.
i) Let $t \in (-1, -b^a].$ Then, by $(*),$ we have $\displaystyle f(t) <\frac{-t^2}{2} \le \frac{-b^{2a}}{2}.$
For $t \ge b^a,$ first see that the function $\displaystyle g(t):=\frac{f(t)}{t}, \ t > -1,$ is decreasing (Exercise!) and thus $g(t) \le g(b^a),$ for all $t \ge b^a.$
On the other hand, we have from $(*)$ that $\displaystyle f(b^a) < -\frac{b^{2a}}{2}+ \frac{b^{3a}}{3} < -\frac{b^{2a}}{2}+\frac{b^{2a}}{3}=\frac{-b^{2a}}{6}$ and so $\displaystyle \frac{f(t)}{t}=g(t) \le g(b^a) =\frac{f(b^a)}{b^a} < \frac{-b^a}{6}.$

ii) By $(*),$ we have $\displaystyle \left|f(t)+\frac{t^2}{2}\right|=\frac{|t|^3}{3(c+1)^3} < \frac{b^{3a}}{3(1-b^a)^3}$ because $|c|< |t| \le b^a. \ \Box$

Problem 2. (Stirling’s formula) Show that $\displaystyle \lim_{n\to\infty} \frac{n!}{n^{n+\frac{1}{2}}e^{-n}}=\sqrt{2\pi}.$

Solution. We have $\displaystyle n!=\int_0^{\infty}x^ne^{-x}dx$ (see Exercise 2 in this post for a hint). Now let’s make the substitution $x=n(1+t)$ to get $\displaystyle n!=n^{n+1}e^{-n}\int_{-1}^{\infty}e^{n(\ln(t+1)-t)}dt.$ Thus $\displaystyle \frac{n!}{n^{n+\frac{1}{2}}e^{-n}}=\sqrt{n} \int_{-1}^{\infty} e^{n(\ln(t+1)-t)}dt. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$
Now choose any real number $a$ such that $\displaystyle \frac{1}{3} < a < \frac{1}{2}$ and let $f(t):=\ln(t+1)-t.$ Then we can rewrite $(1)$ as
$\displaystyle \frac{n!}{n^{n+\frac{1}{2}}e^{-n}}=\sqrt{n} \left(\int_{-1}^{\frac{-1}{n^a}}e^{nf(t)}dt + \int_{\frac{-1}{n^a}}^{\frac{1}{n^a}} e^{nf(t)}dt + \int_{\frac{1}{n^a}}^{\infty} e^{nf(t)}dt \right).$
So we are done if we show that
i) $\displaystyle \lim_{n\to\infty} \sqrt{n} \int_{-1}^{\frac{-1}{n^a}}e^{nf(t)}dt=0,$
ii) $\displaystyle \lim_{n\to\infty} \sqrt{n} \int_{\frac{1}{n^a}}^{\infty} e^{nf(t)}dt=0,$
iii) $\displaystyle \lim_{n\to\infty} \sqrt{n} \int_{\frac{-1}{n^a}}^{\frac{1}{n^a}} e^{nf(t)}dt=\sqrt{2\pi}.$
Proof of i). By part i) of Problem 1, $\displaystyle f(t) < \frac{-1}{2n^{2a}}$ for all $\displaystyle -1 and thus $\displaystyle 0 < \sqrt{n} \int_{-1}^{\frac{-1}{n^a}}e^{nf(t)}dt \le \sqrt{n}\left(1-\frac{1}{n^a}\right)e^{\frac{-1}{2}n^{1-2a}}.$
But since $1-2a > 0,$ we have $\displaystyle \lim_{n\to\infty} \sqrt{n}\left(1-\frac{1}{n^a}\right)e^{\frac{-1}{2}n^{1-2a}}=0$ and the result follows.
Proof of ii). By part i) of Problem 1, we have $\displaystyle f(t) < \frac{-1}{6n^a}t$ for $\displaystyle t \ge \frac{1}{n^a}$ and thus $\displaystyle 0 < \sqrt{n} \int_{\frac{1}{n^a}}^{\infty} e^{nf(t)}dt \le \sqrt{n} \int_{\frac{1}{n^a}}^{\infty} e^{\frac{-1}{6}n^{1-a}t}dt =\frac{6\sqrt{n}}{n^{1-a}}e^{\frac{-1}{6}n^{1-2a}}.$
But since $1-2a > 0,$ we have $\displaystyle \lim_{n\to\infty} \frac{6\sqrt{n}}{n^{1-a}}e^{\frac{-1}{6}n^{1-2a}}=0$ and the result follows.
Proof of iii). By Problem 1, ii), we have
$\displaystyle -\frac{t^2}{2} - \frac{1}{3(n^a-1)^3} < f(t) < -\frac{t^2}{2} + \frac{1}{3(n^a-1)^3},$
for all $\displaystyle \frac{-1}{n^a} \le t \le \frac{1}{n^a}$ and thus
$\displaystyle e^{\frac{-n}{3(n^a-1)^3}}\int_{\frac{-1}{n^a}}^{\frac{1}{n^a}}\sqrt{n} e^{\frac{-nt^2}{2}}dt \le \sqrt{n} \int_{\frac{-1}{n^a}}^{\frac{1}{n^a}} e^{nf(t)}dt \le e^{\frac{n}{3(n^a-1)^3}} \int_{\frac{-1}{n^a}}^{\frac{1}{n^a}} \sqrt{n}e^{\frac{-nt^2}{2}}dt. \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$
But clearly $\displaystyle \lim_{n\to\infty} \frac{n}{3(n^a-1)^3}=0,$ because $1 < 3a,$ and so $(2)$ gives $\displaystyle \lim_{n\to\infty} \sqrt{n} \int_{\frac{-1}{n^a}}^{\frac{1}{n^a}} e^{nf(t)}dt=\lim_{n\to\infty} \int_{\frac{-1}{n^a}}^{\frac{1}{n^a}} \sqrt{n}e^{\frac{-nt^2}{2}}dt=2 \lim_{n\to\infty} \int_0^{\frac{1}{n^a}} \sqrt{n}e^{\frac{-nt^2}{2}}dt. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$
The substitution $\displaystyle \sqrt{\frac{n}{2}}t=s$ now gives
$\displaystyle \int_0^{\frac{1}{n^a}} \sqrt{n}e^{\frac{-nt^2}{2}}dt=\sqrt{2}\int_0^{\sqrt{\frac{1}{2}n^{1-2a}}}e^{-s^2}ds.$
But since $1-2a >0,$ we have $\displaystyle \lim_{n\to\infty} \sqrt{\frac{1}{2}n^{1-2a}}=\infty$ and hence, by this post,
$\displaystyle \lim_{n\to\infty}\int_0^{\frac{1}{n^a}} \sqrt{n}e^{\frac{-nt^2}{2}}dt=\sqrt{2} \int_0^{\infty}e^{-s^2}ds=\sqrt{\frac{\pi}{2}}.$
Therefore, by $(3),$  $\displaystyle \lim_{n\to\infty} \sqrt{n} \int_{\frac{-1}{n^a}}^{\frac{1}{n^a}} e^{nf(t)}dt=2\sqrt{\frac{\pi}{2}}=\sqrt{2\pi}. \ \Box$