10

So we have, $$\begin{align} \lim_{n\to\infty} \sum_{k=1}^{n}\left(\frac{k}{n}\right)^n &= \lim_{n\to\infty} \sum_{j=0}^{n-1}\bigg(\frac{n-j}{n}\bigg)^n \\ &= \lim_{n\to\infty} \bigg(1+\bigg(1-\frac{1}{n}\bigg)^n+...+\bigg(1-\frac{n-1}{n}\bigg)^n\bigg) \\ &= 1+e^{-1}+e^{-2}+... \\ &= \frac{e}{e-1} \end{align},$$

but my problem is going from the second to the third line. As the limit involves both the summand and the sum and I am not sure how this line is "legal", that is what is it that allows is to apply the limit first to each term then to the sum, viz why is that $\lim\limits_{n\to\infty} \sum\limits_{k=0}^{n-1}\bigg(1+\frac{-k}{n}\bigg)^n=\sum\limits_{k=0}^\infty\lim\limits_{n\to\infty}\bigg(1+\frac{-k}{n}\bigg)^n$? Is there a double limit? But can you have a double limit involving the same variable? Is this some special case of Fubini's Theorem, if so I am really struggling to see how? Or maybe it is a Riemann sum? If so so I'm not sure how to show t manipulate it to show that. Any help will be greatly appreciated.

Thanks in advance.

user152874
  • 623
  • You are right to be skeptical of this move. It is definitely something which has to be carefully justified. – Arthur Jan 08 '19 at 07:51
  • 5
    You can justify it by monotone convergence theorem. See this earlier post https://math.stackexchange.com/questions/3003092/to-evaluate-the-limit/3003136#3003136 – Myunghyun Song Jan 08 '19 at 08:00
  • Thanks, this clears it up but if you use the dominated convergence theorem, what is the dominant function? – user152874 Jan 08 '19 at 08:16
  • @user152874 We can take $k\mapsto e^{-k}$ as a dominating function! – Myunghyun Song Jan 08 '19 at 08:54
  • Thanks again, but what is the measure space we are working in to see the sum as a Lebesgue integral? – user152874 Jan 09 '19 at 08:13
  • 1
    @user152874 If you don't add @, I can't get an alarm ... Well, in this case the given measure is a counting measure, usually denoted by $c$. So if $a :\mathbb{N}\to \mathbb{R}$ is a sequence, $\int_\mathbb{N} a_k dc(k) = \sum_{k=1}^\infty a_k$. – Myunghyun Song Jan 10 '19 at 14:19
  • @Song Thanks again and sorry about the "@". Just a quick question about the counting measure in an arbitrary infinite space. In $\mathbb{N}$ for the monotone convergence theorem gives $\int_\mathbb{N} a_k dc(k) = \sum_{k=1}^\infty a_k$, however does this still hold for uncountable spaces? If so how, will we be able to still use the monotone convergence theorem? – user152874 Jan 13 '19 at 08:26
  • Maybe it is worth mentioning that $\int_\mathbb{N} a_k dc(k) := \sum_{k=1}^\infty a_k$ holds by definition, and $\lim_{n\to\infty} \sum_{k} a_{n,k} =\sum_k (\lim_n a_{n,k})$ is a consequence of monotone convergence theorem. Anyway, if we consider an uncountable space $X$, then $\int_X a(x)dc(x) := \sum_{x\in X} a(x)$ still holds for $a(x)\ge 0$. But more subtle point is that $\sum_{x\in X} |a(x)| <\infty$implies that ${x\in X;|;a(x)\neq 0}$ is at most countable set, so the sum is actually taken on a countable set. – Myunghyun Song Jan 13 '19 at 14:03
  • @Song Ok, I think I am a bit confused.My understanding of how the counting measure is just summation is similar to this post https://math.stackexchange.com/a/764089/152874. Could you elaborate or give me a reference on why you say that it holds by definition. Thanks again. – user152874 Jan 13 '19 at 21:37
  • 2
    Possible duplicate of How should I calculate $\lim_{n\rightarrow \infty} \frac{1^n+2^n+3^n+...+n^n}{n^n}$ – StubbornAtom Jan 23 '19 at 14:01
  • Also see https://math.stackexchange.com/questions/927771/what-is-lim-n-to-infty-sum-k-1n-left-frackn-rightn?rq=1. – StubbornAtom Jan 23 '19 at 14:14
  • And https://math.stackexchange.com/questions/164074/how-to-evaluate-lim-limits-n-to-infty-sum-limits-k-1n-fracknn?noredirect=1&lq=1, – StubbornAtom Jun 22 '19 at 16:07

1 Answers1

1

This requires justification.

First use the first $m$ terms with $m$ fixed to see that $\sum\limits_{k=1}^{n} (\frac k n)^{n} \geq 1+(1-\frac 1n)^{n}+\cdots+(1-\frac m n)^{n}$ for $n >m$ which gives $\liminf\sum\limits_{k=1}^{n} (\frac k n)^{n} \geq 1+e^{-1}+e^{-2}+\cdots+e^{-m}$ for each $m$. Now let $m \to \infty$.

Next use the inequality $1-x \leq e^{-x}, x \geq 0$ to prove that for every $n$ we have $ \sum\limits_{k=1}^{n} (\frac k n)^{n} \leq $ RHS. [$(1-\frac j n)^{n} \leq (e^{-j/n})^{n}=e^{-j}$ for each $j$].

Kavi Rama Murthy
  • 311,013