Tannery’s theorem

Here I proved that $\displaystyle \lim_{n\to\infty} \frac{1}{n^n} \sum_{k=1}^n k^n=\frac{e}{e-1}.$ My solution was nice and elementary but not very natural. A more natural solution would begin with the following line

$\displaystyle \frac{1}{n^n} \sum_{k=1}^n k^n=\sum_{k=1}^n \left(\frac{k}{n}\right)^n= \sum_{k=0}^n \left(1-\frac{k}{n}\right)^n$

and it would continue as follows: but $\displaystyle \lim_{n\to\infty} \left(1-\frac{k}{n}\right)^n=e^{-k}$ and so if we can write $\displaystyle \lim_{n\to\infty} \sum_{k=0}^n \left(1-\frac{k}{n}\right)^n=\sum_{k=0}^{\infty} \lim_{n\to\infty} \left(1-\frac{k}{n}\right)^n,$ then the answer would be

$\displaystyle \sum_{k=0}^{\infty}e^{-k}=\frac{1}{1-e^{-1}}=\frac{e}{e-1}.$

Right, but only if we can write it that way. In fact, it turns out that we can!
In general, we have the following result.

Tannery’s theorem. For an integer $k \ge 0,$ let $f_k$ be a function of the integer variable $n.$ Suppose that there exist sequences $\{L_k\}$ and $\{M_k\},$ where each $M_k$ is independent of $n,$ such that $\displaystyle \lim_{n\to\infty} f_k(n)=L_k$ and $|f_k(n)| \le M_k,$ for all $k, n.$ If $\displaystyle \sum_{k=0}^{\infty} M_k$ is convergent, then

$\displaystyle \lim_{n\to\infty} \sum_{k=0}^n f_k(n)=\sum_{k=0}^{\infty} \lim_{n\to\infty}f_k(n)=\sum_{k=0}^{\infty} L_k.$

Proof. Let $\epsilon > 0$ be given. Since $\displaystyle \sum_{k=0}^{\infty} M_k$ is convergent, $\displaystyle \lim_{m\to\infty} \sum_{k=m}^{\infty}M_k=0$ and so

$\displaystyle \sum_{k=m}^{\infty} M_k < \epsilon \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$

if $m$ is large enough. Note that $m$ doesn’t depend on $n$ because $M_k$ doesn’t depend on $n.$ Thus if $m$ is large enough and $n \ge m,$ then

$\displaystyle \left|\sum_{k=m}^n f_k(n) \right| \le \sum_{k=m}^n |f_k(n)| \le \sum_{k=m}^n M_k \le \sum_{k=m}^{\infty} M_k < \epsilon. \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$

Also since $M_k$ doesn’t depend on $\displaystyle n, \ |L_k| =\lim_{n\to\infty} |f_k(n)| \le M_k$ and so, by $(1),$ if $m$ is large enough, then

$\displaystyle \left|\sum_{k=m}^{\infty} L_k \right| \le \sum_{k=m}^{\infty}|L_k| \le \sum_{k=m}^{\infty}M_k < \epsilon. \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$

Therefore, if $m$ is large enough and $n \ge m,$ then $(2)$ and $(3)$ give

$\displaystyle \begin{aligned} \left|\sum_{k=0}^nf_k(n)-\sum_{k=0}^{\infty}L_k\right|=\left|\sum_{k=0}^{m-1}(f_k(n)-L_k)+\sum_{k=m}^n f_k(n) - \sum_{k=m}^{\infty}L_k\right|\end{aligned}$

$\displaystyle \le \sum_{k=0}^{m-1}|f_k(n)-L_k|+ \left|\sum_{k=m}^n f_k(n) \right| + \left|\sum_{k=m}^{\infty} L_k \right|$

$\displaystyle \le \sum_{k=0}^{m-1}|f_k(n)-L_k|+ 2\epsilon. \ \ \ \ \ \ \ \ \ \ \ \ \ (4)$

Finally, since for each $\displaystyle k, \ \lim_{n\to\infty} f_k(n)=L_k$ and $m$ doesn’t depend on $n,$ we have $\displaystyle \lim_{n\to\infty} \sum_{k=0}^{m-1}|f_k(n)-L_k|=0$ and so $\displaystyle \sum_{k=0}^{m-1}|f_k(n)-L_k| < \epsilon$ provided that $n$ is large enough. Hence, by $(4),$

$\displaystyle \left|\sum_{k=0}^nf_k(n)-\sum_{k=0}^{\infty}L_k\right| < 3\epsilon,$

for all $n$ that are large enough. So $\displaystyle \lim_{n\to\infty} \sum_{k=0}^n f_k(n)=\sum_{k=0}^{\infty} L_k,$ by the definition of limit. $\Box$

Example 1. Using Tannery’s theorem, show that $\displaystyle \lim_{n\to\infty} \frac{1}{n^n} \sum_{k=1}^n k^n=\frac{e}{e-1}.$

Solution. We have

$\displaystyle \frac{1}{n^n} \sum_{k=1}^n k^n=\sum_{k=1}^n\left(\frac{k}{n}\right)^n=\sum_{k=0}^n \left(\frac{n-k}{n}\right)^n=\sum_{k=0}^n \left(1-\frac{k}{n}\right)^n.$

Let $\displaystyle f_k(n):=\left(1-\frac{k}{n}\right)^n.$ Then $\displaystyle L_k:=\lim_{n\to\infty}f_k(n)=e^{-k}.$ Also, since $e^x \ge 1+x$ for real numbers $x,$ we also have

$\displaystyle |f_k(n)|=f_k(n)=\left(1-\frac{k}{n}\right)^n \le e^{-k}:=M_k.$

Clearly $M_k$ does not depend on $n$ and $\displaystyle \sum_{k=0}^{\infty}M_k=\sum_{k=0}^{\infty}e^{-k}$ is convergent. Thus, by Tannery’s theorem,

$\displaystyle \lim_{n\to\infty} \frac{1}{n^n} \sum_{k=1}^n k^n=\lim_{n\to\infty} \sum_{k=0}^nf_k(n)=\sum_{k=0}^{\infty} L_k = \sum_{k=0}^{\infty} e^{-k}=\frac{e}{e-1}. \ \Box$

Example 2. Let $a \in \mathbb{R}$ . Using Tannery’s theorem, show that

$\displaystyle \lim_{n\to\infty} \left(1+\frac{a}{n}\right)^n=\sum_{k=0}^{\infty} \frac{a^k}{k!}.$

Solution. By the binomial theorem, $\displaystyle \left(1+\frac{a}{n}\right)^n=\sum_{k=0}^n \binom{n}{k}\frac{a^k}{n^k}.$ Let $\displaystyle f_k(n):=\binom{n}{k}\frac{a^k}{n^k}.$ Then

$\displaystyle f_k(n)=\frac{n!}{n^k(n-k)!} \cdot \frac{a^k}{k!}=\frac{n(n-1) \cdots (n-k+1)}{n^k} \cdot \frac{a^k}{k!}$

and thus $\displaystyle L_k:=\lim_{n\to\infty}f_k(n)=\frac{a^k}{k!}.$ Also, $\displaystyle |f_k(n)| \le \frac{|a|^k}{k!}:=M_k$ because

$\displaystyle \frac{n(n-1) \cdots (n-k+1)}{n^k} \le 1.$

Clearly $M_k$ doesn’t depend on $n$ and $\displaystyle \sum_{k=0}^{\infty}M_k=\sum_{k=0}^{\infty}\frac{|a|^k}{k!}$ is convergent (you may, for example, use the ratio test). Thus, by Tannery’s theorem,

$\displaystyle \lim_{n\to\infty} \left(1+\frac{a}{n}\right)^n=\lim_{n\to\infty} \sum_{k=0}^nf_k(n)=\sum_{k=0}^{\infty} L_k =\sum_{k=0}^{\infty} \frac{a^k}{k!}. \ \Box$

Example 3. Show that $\displaystyle \lim_{n\to\infty } \sum_{k=0}^{n} \frac{\binom{n}{k}}{(k+3)n^k}=e-2.$

Solution. See that, for a fixed $k,$ we have $\displaystyle L_k:=\lim_{n\to\infty} \frac{\binom{n}{k}}{(k+3)n^k}=\frac{1}{k!(k+3)}.$
We also have $\displaystyle \binom{n}{k} \le \frac{n^k}{k!}$ (why?) and thus $\displaystyle \frac{\binom{n}{k}}{n^k(k+3)} \le \frac{1}{k!(k+3)}.$ Since $\displaystyle \sum_{k=0}^{\infty}\frac{1}{k!(k+3)}$ is convergent, we can use Tannery’s theorem and write

$\displaystyle \begin{aligned} \lim_{n\to\infty} \sum_{k=0}^n \frac{\binom{n}{k}}{n^k(k+3)}=\sum_{k=0}^{\infty}L_k=\sum_{k=0}^{\infty} \frac{1}{k!(k+3)}=\int_0^1 x^2e^xdx=e-2. \ \Box \end{aligned}$

Exercise (a stronger version of Tannery’s theorem). Show that Tannery’s theorem still holds if, in the theorem, we replace $\displaystyle \sum_{k=0}^n f_k(n)$ with $\displaystyle \sum_{k=0}^{g(n)}f_k(n),$ where $g: \mathbb{N} \longrightarrow \mathbb{N}$ is any increasing function of $n.$