# Limits involving binomial coefficients (2)

Problem. Show that $\displaystyle \lim_{n\to\infty}\frac{1}{n^2}\ln\left(\prod_{k=0}^n \binom{n}{k}\right)=\frac{1}{2}.$

Solution. We have

\displaystyle \begin{aligned} a_n:=\ln\left(\prod_{k=0}^n \binom{n}{k}\right)=\sum_{k=0}^n (\ln n! - \ln k! - \ln(n-k)!)=(n+1)\ln n! - 2\sum_{k=0}^n \ln k! \\ =(n+1)\sum_{i=1}^n \ln i - 2\sum_{k=1}^n\sum_{i=1}^k \ln i =(n+1)\sum_{i=1}^n \ln i - 2 \sum_{i=1}^n\sum_{k=i}^n\ln i \\ =(n+1)\sum_{i=1}^n \ln i - 2\sum_{i=1}^n(n-i+1)\ln i=\sum_{i=1}^n(2i-n-1)\ln i \\ =\sum_{i=1}^n (2i-n)\ln \left(\frac{i}{n}\right)+\sum_{i=1}^n(2i-n)\ln n - \sum_{i=1}^n \ln i \\ =\sum_{i=1}^n (2i-n)\ln \left(\frac{i}{n}\right)+n\ln n- \ln n!. \end{aligned}

But $\displaystyle \lim_{n\to\infty} \frac{n\ln n}{n^2}=\lim_{n\to\infty} \frac{\ln n!}{n^2}=0$ (because $\displaystyle n! \le n^n$) and so

\displaystyle \begin{aligned} \lim_{n\to\infty} \frac{a_n}{n^2}=\lim_{n\to\infty} \frac{1}{n^2}\sum_{i=1}^n (2i-n)\ln \left(\frac{i}{n}\right)=\lim_{n\to\infty} \frac{1}{n}\sum_{i=1}^n \left(\frac{2i}{n}-1\right)\ln \left(\frac{i}{n}\right) \\ =\int_0^1(2x-1)\ln x \ dx = \frac{1}{2}. \ \Box \end{aligned}

# Limit of integrals (15)

Problem. Show that

i) $\displaystyle u-\frac{u^3}{6} \le \sin u \le u$ for $u \ge 0$ and $\displaystyle \cos u \ge 1- \frac{u^2}{2}$ for all real numbers $u$

ii) $\displaystyle \lim_{x\to+\infty} \int_0^x \sin\left(\frac{1}{x+t}\right) dt=\ln 2$ and $\displaystyle \lim_{x\to+\infty} \int_0^x \cos\left(\frac{1}{x+t}\right) dt=+\infty.$

Solution. i) Let

$\displaystyle f(u):=u-\sin u, \ \ g(u):=\sin u - u+\frac{u^3}{6}, \ \ h(u):=\cos u - 1 + \frac{u^2}{2}.$

The function $f$ is increasing on the real line because $f'(u)=1-\cos u \ge 0.$ So $f(u) \ge f(0)=0$ for $u \ge 0.$
We have $h'(u)=u-\sin u=f(u).$ Therefore $h$ is increasing for $u \ge 0$ and so $h(u) \ge h(0)=0$ for $u \ge 0.$ Since $h$ is an even function, $h(u) \ge 0$ for all real numbers $u.$
We have $\displaystyle g'(u)=\cos u - 1 + \frac{u^2}{2}=h(u).$ So $g$ is increasing for all real numbers $u$ and hence $g(u) \ge g(0)=0$ for $u \ge 0.$

ii) By i), $\displaystyle \frac{1}{x+t} - \frac{1}{6(x+t)^3} \le \sin\left(\frac{1}{x+t}\right) \le \frac{1}{x+t}$ for all $x,t > 0.$ So if $x > 0,$ then

\displaystyle \begin{aligned} \ln 2 - \frac{1}{16x^2}=\int_0^x \frac{dt}{x+t} -\int_0^x \frac{dt}{6(x+t)^3} \le \int_0^x \sin\left(\frac{1}{x+t}\right) dt \le \int_0^x \frac{dt}{x+t}=\ln 2\end{aligned}

and the result follows from the squeeze theorem.
Also, again by i), $\displaystyle \cos\left(\frac{1}{x+t}\right) \ge 1 - \frac{1}{2(x+t)^2}$ for all $x,t > 0.$ So if $x > 0,$ then

$\displaystyle \int_0^x \cos\left(\frac{1}{x+t}\right) dt \ge \int_0^xdt - \int_0^x \frac{dt}{2(x+t)^2}=x-\frac{1}{4x}$

and the result follows. $\Box$

# List of Maclaurin series for some functions

$\displaystyle(x+1)^a=\sum_{n=0}^{\infty}\binom{a}{n}x^n, \ \ \ \ \ a \in \mathbb{R}, \ \ x \in (-1,1)$

Comment. See here!

$\displaystyle \frac{1}{x+1}=\sum_{n=0}^{\infty}(-1)^nx^n, \ \ \ \ \ x \in (-1,1)$

\displaystyle \begin{aligned} \frac{1}{(x+1)^m}=\sum_{n=0}^{\infty}(-1)^n \binom{n+m-1}{n}x^n, \ \ \ \ \ m \ \text{positive integer}, \ x \in (-1,1)\end{aligned}

$\displaystyle \sqrt{x+1}=\sum_{n=0}^{\infty}\frac{(-1)^{n-1}\binom{2n}{n}}{(2n-1)4^n}x^n, \ \ \ \ \ x \in (-1,1)$

$\displaystyle \frac{1}{\sqrt{x+1}}=\sum_{n=0}^{\infty}\frac{(-1)^n\binom{2n}{n}}{4^n}x^n, \ \ \ \ \ x \in (-1,1)$

\displaystyle \begin{aligned} \frac{1}{(x+\cot \theta)^2+1}=\sum_{n=0}^{\infty}(-1)^n \sin((n+1)\theta)(\sin \theta)^{n+1}x^n, \ \ \ \ \ \theta \in (0, \pi), \ x \in (-\csc \theta, \csc \theta) \end{aligned}

Comment. See here!

$\displaystyle \frac{1}{1-x-x^2}=\sum_{n=0}^{\infty}F_nx^n, \ \ \ \ \ x \in \left(\frac{1-\sqrt{5}}{2}, \frac{\sqrt{5}-1}{2}\right)$

Comment. See here and here!

$\displaystyle \ln(x+1)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}x^n, \ \ \ \ \ x \in (-1,1]$

$\displaystyle (\ln(x+1))^2=2\sum_{n=2}^{\infty} \left(\frac{(-1)^n}{n}\sum_{k=1}^{n-1} \frac{1}{k}\right)x^n, \ \ \ \ \ x \in (-1,1]$

Comment. It follows from $\displaystyle (\ln(x+1))^2=2\int_0^x \frac{\ln(t+1)}{t+1} \ dt.$

$\displaystyle e^{e^x} = \sum_{n=0}^{\infty} \frac{b_ne}{n!}x^n, \ \ \ \ \ x \in \mathbb{R}$

Comment. See here! (Example 2)

$\displaystyle \frac{x}{e^x-1}=\sum_{n=0}^{\infty} \frac{B_n}{n!}x^n, \ \ \ \ \ x \in (-2\pi,2\pi)$

Comment. See here!

$\displaystyle \sinh x = \frac{e^x-e^{-x}}{2}=\sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}, \ \ \ \ \ x \in \mathbb{R}$

$\displaystyle \cosh x = \frac{e^x+e^{-x}}{2}=\sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}, \ \ \ \ \ x \in \mathbb{R}$

$\displaystyle \tanh x = \sum_{n=1}^{\infty} \frac{4^n(4^n-1)B_{2n}}{(2n)!}x^{2n-1}, \ \ \ \ \ x \in \left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$

Comment. The proof is similar to the one given for the Maclurin series of $\tan x.$ Note that $\displaystyle \coth x$ has no Maclaurin series expansion because it is not even defined at $x=0.$

$\displaystyle \sinh^{-1}x=\sum_{n=0}^{\infty} \frac{(-1)^n\binom{2n}{n}}{(2n+1)4^n}x^{2n+1}, \ \ \ \ \ x \in (-1,1)$

$\displaystyle \tanh^{-1} x=\frac{1}{2}\ln\left(\frac{1+x}{1-x}\right)=\sum_{n=1}^{\infty}\frac{x^{2n-1}}{2n-1}, \ \ \ \ \ x \in (-1,1)$

$\displaystyle \sin x = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)!}x^{2n+1}, \ \ \ \ \ x \in \mathbb{R}$

$\displaystyle \cos x = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n)!}x^{2n}, \ \ \ \ \ x \in \mathbb{R}$

$\displaystyle \tan x = \sum_{n=1}^{\infty} (-1)^{n-1}\frac{4^n(4^n-1)B_{2n}}{(2n)!}x^{2n-1}, \ \ \ \ \ x \in \left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$

Comment.  See here! Note that $\displaystyle \cot x$ has no Maclaurin series expansion because it is not even defined at $x=0.$

$\displaystyle \sin^{-1}x = \sum_{n=0}^{\infty} \frac{\binom{2n}{n}}{(2n+1)4^n}x^{2n+1}, \ \ \ \ \ x \in [-1,1]$

Comment. See here! Note that the Maclaurin series for $\displaystyle \cos^{-1}x$ now follows from the identity $\displaystyle \cos^{-1}x = \frac{\pi}{2}-\sin^{-1}x.$

$\displaystyle (\sin^{-1}x)^2=\sum_{n=1}^{\infty} \frac{2^{2n-1}}{n^2\binom{2n}{n}}x^{2n}, \ \ \ \ \ x \in [-1,1]$

Comment: see here!

$\displaystyle \tan^{-1}x = \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}x^{2n+1}, \ \ \ \ \ x \in [-1,1]$

Comment. Note that the Maclaurin series for $\displaystyle \cot^{-1}x$ now follows from the identity $\displaystyle \cot^{-1}x = \frac{\pi}{2}-\tan^{-1}x.$

$\displaystyle (\tan^{-1}x)^2=\sum_{n=1}^{\infty}\left(\frac{(-1)^{n-1}}{n}\sum_{k=1}^n \frac{1}{2k-1}\right)x^{2n}, \ \ \ \ \ x \in [-1,1]$

Comment. It follows from $\displaystyle (\tan^{-1}x)^2=2\int_0^x \frac{\tan^{-1}t}{t^2+1} \ dt.$

# Integral of x^k*ln(sin(x))

Notation. Throughout this post, $\displaystyle I_k:=\int_0^{\frac{\pi}{2}}x^k\ln(\sin x) \ dx,$ where $k$ is an integer.

We showed here that $\displaystyle I_0=\int_0^{\frac{\pi}{2}} \ln(\sin x) \ dx=-\frac{\pi}{2}\ln 2,$ which was done quite easily. Now, let’s ask a more interesting question: given an integer $k \ge 0,$ how can we evaluate $\displaystyle I_k=\int_0^{\frac{\pi}{2}}x^k\ln(\sin x) \ dx ?$

What is certain is that the method we used to evaluate $I_k$ for $k=0$ is absolutely useless for $k > 0.$ We need a more powerful tool and we have already seen that tool in this post. As I mentioned in the Remark in that post, we have

$\displaystyle \ln(\sin x)= -\ln 2 -\sum_{n=1}^{\infty} \frac{\cos(2nx)}{n}, \ \ \ \ \ x \in (0,\pi) \ \ \ \ \ \ \ \ \ \ \ \ (1)$

and thus

$\displaystyle I_k= -\frac{\ln 2}{k+1}\left(\frac{\pi}{2}\right)^{k+1} - \sum_{n=1}^{\infty}\frac{1}{n}\int_0^{\frac{\pi}{2}} x^k\cos(2nx) \ dx. \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$

Of course, in order to get $(2),$ we integrated the infinite series $\displaystyle \sum_{n=1}^{\infty} \frac{\cos(2nx)}{n}$ in $(1)$ term by term. But can we do that? Yes, we can but I won’t explain the reason here.

Problem. Given integers $k \ge 0, \ n \ge 1,$ let $\displaystyle J_k:=\int_0^{\frac{\pi}{2}}x^k \cos(2nx) \ dx,$ Show that

i) $\displaystyle J_0=0, \ \ J_1=\frac{(-1)^n-1}{4n^2}$ and

$\displaystyle J_k=\frac{(-1)^nk}{4n^2}\left(\frac{\pi}{2}\right)^{k-1}-\frac{k(k-1)}{4n^2}J_{k-2}, \ \ k \ge 2.$

ii) $\displaystyle I_0=-\frac{\pi}{2}\ln 2, \ \ I_1=-\frac{\pi^2}{8}\ln 2+\frac{7}{16}\zeta(3)$ and

\displaystyle \begin{aligned} I_k=-\frac{\pi^{k+1}}{(k+1)2^{k+1}}\ln 2 + \frac{3k\pi^{k-1}}{2^{k+3}}\zeta(3)+\frac{k(k-1)}{4}\sum_{n=1}^{\infty}\frac{1}{n^3}J_{k-2}, \ \ k \ge 2.\end{aligned}

Solution. i) Integration by parts.

ii) It follows easily from i), $(2)$ and the first part of Problem 1 in this post! $\Box$

Example 1. Evaluate $\displaystyle \int_0^{\frac{\pi}{2}}x^2\ln(\sin x) \ dx$ and $\displaystyle \int_0^{\frac{\pi}{2}}x^3\ln(\sin x) \ dx.$

Solution. By the second part of the above problem,

$\displaystyle I_2=-\frac{\pi^3}{24}\ln 2 + \frac{3\pi}{16}\zeta(3)+\frac{1}{2}\sum_{n=1}^{\infty} \frac{1}{n^3}J_0.$

By the first part of the above problem, $J_0=0$ and so $\displaystyle I_2=-\frac{\pi^3}{24}\ln 2 + \frac{3\pi}{16}\zeta(3).$
Again, by the second part of the above problem,

$\displaystyle I_3= - \frac{\pi^4}{64}\ln 2 + \frac{9\pi^2}{64}\zeta(3)+\frac{3}{2}\sum_{n=1}^{\infty} \frac{1}{n^3}J_1. \ \ \ \ \ \ \ \ \ \ \ \ (3)$

By the first part of the above problem, $\displaystyle J_1=\frac{(-1)^n-1}{4n^2}$ and so by the first part of Problem 1 in this post, we have

\displaystyle \begin{aligned}\sum_{n=1}^{\infty} \frac{1}{n^3}J_1=\frac{1}{4}\left(\sum_{n=1}^{\infty} \frac{(-1)^n}{n^5}-\sum_{n=1}^{\infty} \frac{1}{n^5}\right)=-\frac{1}{4}((1-2^{-4})\zeta(5)+\zeta(5))=-\frac{31}{64}\zeta(5).\end{aligned}

Thus, by $\displaystyle (3), \ I_3= - \frac{\pi^4}{64}\ln 2 + \frac{9\pi^2}{64}\zeta(3)-\frac{93}{128}\zeta(5). \ \Box$

Example 2. Evaluate $\displaystyle \int_0^{\pi}x^2\ln(\sin x) \ dx.$

Solution. We have

\displaystyle \begin{aligned} \int_0^{\pi}x^2\ln(\sin x) \ dx= \int_0^{\frac{\pi}{2}}x^2\ln(\sin x) \ dx + \int_{\frac{\pi}{2}}^{\pi}x^2\ln(\sin x) \ dx \\ =I_2 +\int_0^{\frac{\pi}{2}}(\pi-x)^2\ln(\sin x) \ dx \\ =\pi^2I_0-2\pi I_1+2I_2. \end{aligned}

The values of $I_0,I_1$ are given in the second part of the above problem and we found $I_2$ in the above example. $\Box$

# Limit of integrals (14)

Problem. Show that

i) $\displaystyle \lim_{n\to\infty} \int_0^1 \ln(1+x+x^n) \ dx =2\ln 2 -1$

ii) $\displaystyle \lim_{n\to\infty} \int_0^1 \ln(1-x+x^n) \ dx =-1.$

Solution. i) Let $\displaystyle I_n:=\int_0^1 \ln(1+x+x^n) \ dx.$ It is clear that

$\displaystyle I_n \ge \int_0^1 \ln(1+x) \ dx = 2\ln 2 - 1. \ \ \ \ \ \ \ \ \ \ \ (1)$

On the other hand, since $\ln(1+t) \le t$ for $t > -1,$ we have

\displaystyle \begin{aligned}I_n=\int_0^1 \ln(1+x) \ dx + \int_0^1 \ln\left(1+\frac{x^n}{x+1}\right)dx \le 2\ln 2 - 1 + \int_0^1 \frac{x^n}{x+1} \ dx \\ \le 2\ln 2 - 1 +\int_0^1 x^n \ dx = 2\ln 2 - 1 + \frac{1}{n+1}.\end{aligned} \ \ \ \ \ \ \ \ \ \ (2)

The result now follows from $(1),(2)$ and the squeeze theorem.

ii) Let $\displaystyle J_n:=\int_0^1 \ln(1-x+x^n) \ dx.$ It is clear that

$\displaystyle J_n \ge \int_0^1 \ln(1-x) \ dx = -1 \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$

On the other hand, since $\ln(1+t) \le \sqrt{t}$ for $t \ge 0$ (prove it!), we have

\displaystyle \begin{aligned} J_n=\int_0^1 \ln(1-x) \ dx + \int_0^1 \ln\left(1+\frac{x^n}{1-x}\right)dx \le -1 + \int_0^1 \sqrt{\frac{x^n}{1-x}} \ dx \\ =-1 + 2\int_0^{\frac{\pi}{2}} \sin^{n+1}\theta \ d\theta. \end{aligned} \ \ \ \ \ \ \ \ \ \ (4)

Notice that we got the last equality in $(4)$ by making the substitution $x = \sin^2\theta.$ The result now follows from $(3),(4),$ the Example in this post and the squeeze theorem. $\Box$

# Limit of integrals (13)

Problem. Let $\displaystyle I_n:=\int_0^1 \frac{dx}{1+x+x^2+ \cdots + x^n}.$ Show that

i) $\displaystyle \lim_{n\to\infty}I_n=\frac{1}{2}$

ii) $\displaystyle \lim_{n\to\infty}n^2\left(I_n - \frac{1}{2}\right)=\frac{\pi^2}{6}.$

Solution. We have

\displaystyle \begin{aligned}I_{n-1}=\int_0^1 \frac{1-x}{1-x^n} \ dx=\int_0^1(1-x)\sum_{k=0}^{\infty}x^{kn}=\int_0^1 \left(\sum_{k=0}^{\infty}x^{kn}-\sum_{k=0}^{\infty}x^{kn+1}\right) dx \\ =\sum_{k=0}^{\infty}\int_0^1 \left(x^{kn}-x^{kn+1}\right)dx=\sum_{k=0}^{\infty}\left(\frac{1}{kn+1}-\frac{1}{kn+2}\right) \\ =\sum_{k=0}^{\infty} \frac{1}{(kn+1)(kn+2)}=\frac{1}{2}+\sum_{k=1}^{\infty} \frac{1}{(kn+1)(kn+2)}.\end{aligned} \ \ \ \ \ \ (*)

But it’s clear that $\displaystyle k^2n^2 < (kn+1)(kn+2) < k^2(n+2)^2,$ for $k \ge 1,$ and so, by $(*),$

$\displaystyle \frac{1}{(n+2)^2} \sum_{k=1}^{\infty}\frac{1}{k^2} < I_{n-1} -\frac{1}{2} < \frac{1}{n^2} \sum_{k=1}^{\infty}\frac{1}{k^2}.$

The results now follow because $\displaystyle \zeta(2)=\sum_{k=1}^{\infty}\frac{1}{k^2}=\frac{\pi^2}{6}$ (see here for an elementary proof of $\displaystyle\sum_{k=1}^{\infty} \frac{1}{k^2}=\frac{\pi^2}{6}$ and see here for basic facts about the zeta function $\zeta$). $\Box$

# Limit of integrals (12)

Problem. Show that $\displaystyle \lim_{n\to\infty} n! \int_0^1 \frac{dx}{(x+1)(x+2) \cdots (x+n)}=0.$

Solution. We know from Bernoulli’s inequality that $\displaystyle 1+\frac{x}{k} \ge (1+x)^{\frac{1}{k}}$ for all integers $k \ge 1.$ So if we let $\displaystyle H_n:=\sum_{k=1}^n \frac{1}{k},$ then we will have

\displaystyle \begin{aligned} 0 \le n! \int_0^1 \frac{dx}{(x+1)(x+2) \cdots (x+n)}=\int_0^1 \frac{dx}{(1+x)(1+\frac{x}{2}) \cdots (1+\frac{x}{n})} \\ \le \int_0^1 \frac{dx}{(1+x)(1+x)^{\frac{1}{2}} \cdots (1+x)^{\frac{1}{n}}} =\int_0^1 (1+x)^{-H_n} \ dx \\ =\frac{1-2^{1-H_n}}{H_n-1} \end{aligned}

and the result follows because $\displaystyle \lim_{n\to\infty} H_n=\sum_{k=1}^{\infty}\frac{1}{k}=\infty. \ \Box$

Definition. The number $\displaystyle H_n:=\sum_{k=1}^n \frac{1}{k}$ is called the $n$-th harmonic number.

Remark 1. The upper bound that we found for $\displaystyle n! \int_0^1 \frac{dx}{(x+1)(x+2) \cdots (x+n)},$ i.e. $\displaystyle \frac{1-2^{1-H_n}}{H_n-1},$ is not a lousy one. For example, for $n=5,$ we have

$\displaystyle 5! \int_0^1 \frac{dx}{(x+1)(x+2)(x+3)(x+4)(x+5)} \approx 0.435$

and $\displaystyle \frac{1-2^{1-H_5}}{H_5-1} \approx 0.459$

Remark 2. Notice that $\displaystyle \frac{n!}{(x+1)(x+2) \cdots (x+n)}=\binom{x+n}{n}^{-1}$ (see this post!).

# The Riemann zeta function

The Riemann zeta function $\zeta$ is defined by $\displaystyle \zeta(\alpha)=\sum_{n=1}^{\infty} \frac{1}{n^{\alpha}},$ where $\alpha > 1$ is a real number. Note that $\zeta(\alpha)$ is well-defined because the series  $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{\alpha}},$ is convergent for $\alpha > 1.$ It is known that if $k \ge 1$ is an integer, then

$\displaystyle \zeta(2k)=(-1)^{k+1} \frac{2^{2k-1}\pi^{2k}B_{2k}}{(2k)!},$

where $B_{2k}$ are Bernoulli numbers. So, for example,

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^2}=\zeta(2)=\frac{2\pi^2B_2}{2!}=\frac{\pi^2}{6}, \ \ \ \sum_{n=1}^{\infty} \frac{1}{n^4}=\zeta(4)=\frac{-8\pi^4B_4}{4!}=\frac{\pi^4}{90}.$

Here we saw a very elementary proof of $\displaystyle \zeta(2)=\frac{\pi^2}{6}.$
There is no closed form for $\zeta(2k+1)$ but the number $\zeta(3) \approx 1.202$ is lucky enough to have a name; it’s called Apéry’s constant.

Problem 1. Show that

i) $\displaystyle \sum_{n=1}^{\infty} \frac{1}{(2n-1)^{\alpha}}=(1-2^{-\alpha})\zeta(\alpha)$ and $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{\alpha}}=(1-2^{1-\alpha})\zeta(\alpha),$

ii) $\displaystyle 1 +2^{-\alpha} < \zeta(\alpha) < \frac{1}{1-2^{1-\alpha}},$

iii) $\displaystyle \lim_{\alpha\to\infty} \zeta(\alpha)=1$ and $\displaystyle \lim_{\alpha\to\infty} (\zeta(\alpha)-1)^{\frac{1}{\alpha}}=\frac{1}{2}.$

Solution. i) We have $\displaystyle \sum_{n=1}^{\infty} \frac{1}{(2n-1)^{\alpha}}=\sum_{n=1}^{\infty} \frac{1}{n^{\alpha}}-\sum_{n=1}^{\infty} \frac{1}{(2n)^{\alpha}}=\zeta(\alpha)-2^{-\alpha}\zeta(\alpha).$
And for the other one

\displaystyle \begin{aligned} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{\alpha}}=\sum_{n=1}^{\infty} \frac{1}{(2n-1)^{\alpha}}-\sum_{n=1}^{\infty} \frac{1}{(2n)^{\alpha}}=(1-2^{-\alpha})\zeta(\alpha)-2^{-\alpha}\zeta(\alpha).\end{aligned}

ii) It’s clear that $\displaystyle \zeta(\alpha)=\sum_{n=1}^{\infty} \frac{!}{n^{\alpha}} > 1+ \frac{1}{2^{\alpha}}.$ We also have

\displaystyle \begin{aligned} \zeta(\alpha)=1+\sum_{n=1}^{\infty}\left(\frac{1}{(2n)^{\alpha}}+\frac{1}{(2n+1)^{\alpha}}\right) < 1+ \sum_{n=1}^{\infty} \frac{2}{(2n)^{\alpha}}=1+2^{1-n}\zeta(\alpha) \end{aligned}

and thus $\displaystyle \zeta(\alpha) < \frac{1}{1-2^{1-\alpha}}.$

iii) Trivial results of ii) and the squeeze theorem. $\Box$

Problem 2. Let $m > 0$ be an integer. Show that

i) $\displaystyle \int_0^{\infty} \frac{x^me^{-x}}{1-e^{-x}} \ dx = m! \zeta(m+1)$

ii) $\displaystyle \int_0^{\infty} \frac{x^me^{-x}}{1-e^{-2x}} \ dx = m!(1-2^{-m-1})\zeta(m+1).$

Solution. The solution is based on the fact that $\displaystyle \int_0^{\infty}x^me^{-ax} dx = \frac{m!}{a^{m+1}}$ for integers $m \ge 0$ and real numbers $a>0$ (see Exercise 2 in this post!).
i) We have

\displaystyle \begin{aligned} \int_0^{\infty} \frac{x^me^{-x}}{1-e^{-x}} \ dx=\int_0^{\infty}x^me^{-x}\sum_{k=0}^{\infty}e^{-kx}dx = \sum_{k=0}^{\infty}\int_0^{\infty}x^me^{-(k+1)x}dx \\ = \sum_{k=0}^{\infty} \frac{m!}{(k+1)^{m+1}}=m!\zeta(m+1).\end{aligned}

ii) We have

\displaystyle \begin{aligned} \int_0^{\infty} \frac{x^me^{-x}}{1-e^{-2x}} \ dx=\int_0^{\infty}x^me^{-x}\sum_{k=0}^{\infty}e^{-2kx}dx = \sum_{k=0}^{\infty}\int_0^{\infty}x^me^{-(2k+1)x}dx = \sum_{k=0}^{\infty} \frac{m!}{(2k+1)^{m+1}}\end{aligned}

and the result follows from Problem 1, i). $\Box$

Problem 3. Show that

i) $\displaystyle \sum_{k=2}^{\infty}(\zeta(k)-1)=1$

ii) $\displaystyle \sum_{k=2}^{\infty} \frac{\zeta(k)-1}{k}=1-\gamma,$ where $\gamma$ is the Euler’s constant.

Solution. i) We have

$\displaystyle \sum_{k=2}^{\infty}(\zeta(k)-1)=\sum_{k=2}^{\infty}\sum_{n=2}^{\infty} \frac{1}{n^k}=\sum_{n=2}^{\infty}\sum_{k=2}^{\infty}\frac{1}{n^k}=\sum_{n=2}^{\infty}\frac{1}{n(n-1)}=1.$

Note that $\displaystyle \sum_{k=2}^{\infty}\frac{1}{n^k}$ is a geometric series and $\displaystyle \sum_{n=2}^{\infty}\frac{1}{n(n-1)}=1$ is a telescoping series.

ii) We have

$\displaystyle \sum_{k=2}^{\infty} \frac{\zeta(k)-1}{k}=\sum_{k=2}^{\infty}\sum_{n=2}^{\infty} \frac{1}{kn^k}=\sum_{n=2}^{\infty}\sum_{k=2}^{\infty} \frac{1}{kn^k}=-\sum_{n=2}^{\infty}\left(\ln\left(1-\frac{1}{n}\right)+\frac{1}{n}\right). \ \ \ \ \ \ \ \ (*)$

Note that the last equality in $(1)$ follows if in the Maclaurin series expansion of $\displaystyle \ln(1-x),$ we put $\displaystyle x = \frac{1}{n}, \ n \ge 2.$
Now for integer $N \ge 2$ we have

\displaystyle \begin{aligned} \sum_{n=2}^N\left(\ln\left(1-\frac{1}{n}\right)+\frac{1}{n}\right)=\sum_{n=2}^N\left(\ln(n-1)-\ln n\right) + \sum_{n=2}^N \frac{1}{n}=-\ln N + \sum_{n=1}^{\infty}\frac{1}{n}-1\end{aligned}

and thus

\displaystyle \begin{aligned} \sum_{n=2}^{\infty}\left(\ln\left(1-\frac{1}{n}\right)+\frac{1}{n}\right)=\lim_{N\to\infty}\sum_{n=2}^N\left(\ln\left(1-\frac{1}{n}\right)+\frac{1}{n}\right) =-1+\lim_{N\to\infty}\left(\sum_{n=1}^{\infty}\frac{1}{n}-\ln N\right) \\ =-1+\gamma.\end{aligned}

The result now follows from $(*). \ \Box$

Exercise 1. Evaluate $\displaystyle \zeta(6)$ using the general formula for $\zeta(2k)$ given at the beginning of this post and compare the result with the bounds for $\zeta(6)$ given in Problem 1, ii).
Hint. $\displaystyle B_6=\frac{1}{42}$ (see Exercise 3 in this post!).

Exercise 2. This exercises generalizes the results given in Problem 2. For the definition of the gamma function $\Gamma(x)$ see this post.
Let $\alpha > 0$ be a real number. Show that

i) $\displaystyle \int_0^{\infty} \frac{x^{\alpha}e^{-x}}{1-e^{-x}} \ dx =\Gamma(\alpha+1) \zeta(\alpha+1)$

ii) $\displaystyle \int_0^{\infty} \frac{x^{\alpha}e^{-x}}{1-e^{-2x}} \ dx = \Gamma(\alpha+1)(1-2^{-\alpha-1})\zeta(\alpha+1).$

# A useful little remark on lim (1+a_n)^n

Problem. Let $\{a_n\}$ be a sequence and suppose that $\displaystyle \lim_{n\to\infty} na_n=a \in (-\infty, +\infty].$ Show that $\displaystyle \lim_{n\to\infty} (1+a_n)^n=e^a.$

Solution. We consider two cases.
Case 1: $a=+\infty.$ So for any number $M > 0,$ we have $na_n > M$ if $n$ is large enough. Thus $\displaystyle (1+a_n)^n > \left(1+\frac{M}{n}\right)^n$ and therefore, because $\displaystyle \lim_{n\to\infty} \left(1+\frac{M}{n}\right)^n=e^M$ and $\displaystyle \lim_{M\to+\infty}e^M=+\infty,$ we have $\displaystyle \lim_{n\to\infty} (1+a_n)^n=+\infty.$

Case 2: $a \in (-\infty,+\infty).$ So given $\epsilon > 0,$ there exists $N > 0$ such that $\displaystyle |na_n - a| < \epsilon$ whenever $n \ge N.$ Thus $\displaystyle 1+\frac{a-\epsilon}{n} < 1+a_n < 1+\frac{a+\epsilon}{n}.$ Note that since $a$ is finite, $\displaystyle 1+\frac{a-\epsilon}{n} \ge 0$ if $n$ is large enough and so

$\displaystyle \left(1+\frac{a-\epsilon}{n}\right)^n < (1+a_n)^n < \left(1+\frac{a+\epsilon}{n}\right)^n.$

Thus $\displaystyle \lim_{n\to\infty}(1+a_n)^n =e^a$ because $\displaystyle \lim_{n\to\infty} \left(1+\frac{a \pm \epsilon}{n}\right)^n=e^{a \pm \epsilon}$ and $\displaystyle \lim_{\epsilon\to0^{+}} e^{a \pm \epsilon}=e^a. \ \Box$

Example 1. Given an integer $k \ge 0,$ we have $\displaystyle \lim_{n\to\infty}n\sum_{i=0}^k \frac{1}{n+i}=k+1$ and thus, by the above problem, $\displaystyle \lim_{n\to\infty} \left(1+\sum_{i=0}^k\frac{1}{n+i}\right)^n=e^{k+1}.$

Example 2. Show that $\displaystyle \lim_{n\to\infty} \left(1+\left|\ln 2 + \sum_{k=1}^n \frac{(-1)^k}{k} \right|\right)^n=\sqrt{e}.$

Solution. We have \displaystyle \begin{aligned}\frac{1}{1+x}=\sum_{k=0}^{n-1} (-1)^k x^k+(-1)^n \frac{x^n}{1+x} \end{aligned} and so integrating over the interval $[0,1]$ gives

$\displaystyle \ln 2 = \sum_{k=1}^n \frac{(-1)^{k-1}}{k}+(-1)^n \int_0^1 \frac{x^n}{1+x} \ dx.$

Thus $\displaystyle \left|\ln 2 + \sum_{k=1}^n \frac{(-1)^k}{k} \right|=\int_0^1 \frac{x^n}{1+x} \ dx$ and hence

$\displaystyle \lim_{n\to\infty} n \left|\ln 2 + \sum_{k=1}^n \frac{(-1)^k}{k} \right|=\lim_{n\to\infty} \int_0^1 \frac{x^n}{1+x} \ dx = \frac{1}{2},$

by the second part of the problem in this post. The result now follows from the above problem. $\Box$

Exercise. Show that the result given in the above problem
i) is true if $a=-\infty$ and the sequence $\{1+a_n\}$ is eventually non-negative.
ii) may or may not be true if $\{1+a_n\}$ is not eventually non-negative.

# Integral of cos(ax)e^(-x^2) and integral of sin(ax)e^(-x^2)

Problem 1. i) Show that

i) $\displaystyle \int_0^{\infty}x^{2n}e^{-x^2}dx=\frac{(2n)! \sqrt{\pi}}{2^{2n+1}n!}$ for all integers $n \ge 0.$

ii) $\displaystyle \int_0^{\infty}\cos(ax) e^{-x^2} dx = \frac{\sqrt{\pi}}{2}e^{-\frac{a^2}{4}}$ for all real numbers $a.$

Solution. i) Let $\displaystyle I_n:=\int_0^{\infty}x^{2n}e^{-x^2}dx.$ We showed here that $\displaystyle I_0=\int_0^{\infty} e^{-x^2}=\frac{\sqrt{\pi}}{2}.$ Now, in $I_n,$ we apply integration by parts with $\displaystyle x^{2n-1}=u$ and $\displaystyle xe^{-x^2}dx=dv$ to get $\displaystyle I_n=\frac{2n-1}{2}I_{n-1}$ and thus

\displaystyle \begin{aligned} I_n=\frac{2n-1}{2} \cdot \frac{2n-3}{2}I_{n-2} = \cdots = \frac{2n-1}{2} \cdot \frac{2n-3}{2} \cdots \frac{1}{2} \cdot I_0 \\ =\frac{(2n)!}{2^n(2n) (2n-2) \cdots 2}I_0 =\frac{(2n)! \sqrt{\pi}}{2^{2n+1}n!}. \end{aligned}

ii) Using i) and the Maclaurin series of $\cos(ax),$ we have

\displaystyle \begin{aligned} \int_0^{\infty}\cos(ax) e^{-x^2}\ dx= \int_0^{\infty}\sum_{n=0}^{\infty}(-1)^n\frac{(ax)^{2n}}{(2n)!}e^{-x^2}dx=\sum_{n=0}^{\infty} \frac{(-a^2)^n}{(2n)!} \int_0^{\infty}x^{2n}e^{-x^2}dx \\ =\sum_{n=0}^{\infty} \frac{(-a^2)^n}{(2n)!} \cdot \frac{(2n)! \sqrt{\pi}}{2^{2n+1}n!} =\frac{\sqrt{\pi}}{2}\sum_{n=0}^{\infty}\frac{(\frac{-a^2}{4})^n}{n!} \\ =\frac{\sqrt{\pi}}{2}e^{\frac{-a^2}{4}}. \ \Box \end{aligned}

Unlike $\displaystyle \int_0^{\infty}\cos(ax) e^{-x^2} dx,$ there’s no closed form for $\displaystyle \int_0^{\infty}\sin(ax) e^{-x^2} dx$ but still there’s something nice and non-trivial that we can prove about the integral, as the following problem shows.

Problem 2. i) Show that

i) $\displaystyle \int_0^{\infty}x^{2n+1}e^{-x^2}dx=\frac{n!}{2}$ for all integers $n \ge 0.$

ii) $\displaystyle \int_0^{\infty}\sin(ax) e^{-x^2} dx = e^{\frac{-a^2}{4}} \int_0^{\frac{a}{2}}e^{x^2}dx$ for all real numbers $a.$

Solution. i) Let $\displaystyle J_n:=\int_0^{\infty}x^{2n+1}e^{-x^2}dx.$ See that $\displaystyle J_0=\int_0^{\infty}xe^{-x^2}dx=\frac{1}{2}.$ Integration by parts with $x^{2n}=u$ and $xe^{-x^2}dx=dv$ gives

$\displaystyle J_n=nJ_{n-1}=n(n-1)J_{n-2}= \cdots = n(n-1) \cdots 2 \cdot 1 \cdot J_0=\frac{n!}{2}.$

ii) Using i) and the Maclaurin series of $\sin(ax),$ we have

\displaystyle \begin{aligned} \int_0^{\infty}\sin(ax) e^{-x^2} dx= \int_0^{\infty}\sum_{n=0}^{\infty}(-1)^n\frac{(ax)^{2n+1}}{(2n+1)!}e^{-x^2}dx \\ =a\sum_{n=0}^{\infty} \frac{(-a^2)^n}{(2n+1)!} \int_0^{\infty}x^{2n+1}e^{-x^2}dx =\frac{a}{2}\sum_{n=0}^{\infty} \frac{(-a^2)^{n}n!}{(2n+1)!} \\ = \frac{a}{2}\sum_{n=0}^{\infty} \frac{(\frac{-a^2}{4})^n4^n}{(2n+1)\binom{2n}{n}n!}. \end{aligned} \ \ \ \ \ \ \ \ \ \ \ \ (*)

But we showed here that $\displaystyle \int_0^{\frac{\pi}{2}} \sin^{2n+1}x \ dx = \frac{4^n}{(2n+1)\binom{2n}{n}}.$ Thus the substitution $x=\cos t$ gives

$\displaystyle \int_0^1 (1-x^2)^ndx = \int_0^{\frac{\pi}{2}} \sin^{2n+1}t \ dx=\frac{4^n}{(2n+1)\binom{2n}{n}}.$

So, by $(*),$

\displaystyle \begin{aligned} \int_0^{\infty}\sin(ax) e^{-x^2} dx= \frac{a}{2}\sum_{n=0}^{\infty} \frac{(\frac{-a^2}{4})^n}{n!}\int_0^1(1-x^2)^n dx=\frac{a}{2}\int_0^1 \sum_{n=0}^{\infty} \frac{(\frac{-a^2(1-x^2)}{4})^n}{n!}dx \\ =\frac{a}{2} \int_0^1 e^{\frac{-a^2(1-x^2)}{4}}dx =\frac{a}{2}e^{\frac{-a^2}{4}}\int_0^1 e^{\frac{a^2x^2}{4}}dx \\ = e^{\frac{-a^2}{4}} \int_0^{\frac{a}{2}}e^{x^2}dx. \ \Box \end{aligned}

Exercise 1. In Problem 2, ii), we showed that $\displaystyle \frac{1}{2}\sum_{n=0}^{\infty}(-1)^n\frac{a^{2n+1}n!}{(2n+1)!}= e^{\frac{-a^2}{4}} \int_0^{\frac{a}{2}}e^{x^2}dx,$ for all real numbers $a.$ Here’s another way to prove it. Let

$\displaystyle f(a):=\frac{1}{2}e^{\frac{a^2}{4}}\sum_{n=0}^{\infty}(-1)^n\frac{a^{2n+1}n!}{(2n+1)!}- \int_0^{\frac{a}{2}}e^{x^2}dx.$

Show that $f'(a)=0$ and conclude that $f(a)=0.$

Exercise 2. Evaluate $\displaystyle \lim_{a\to\infty} \int_0^{\infty} \sin(ax)e^{-x^2}dx$ and $\displaystyle \lim_{a\to\infty} a \int_0^{\infty} \sin(ax)e^{-x^2}dx.$