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.


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