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I am trying to evaluate this limit without using the closed form expression for the sum of natural numbers raised to $k$th power. $$\lim_{n \to \infty} \dfrac{ 1^n +2^n+\cdots +n^n}{n^n}$$

So far I have tried l'Hôpital which complicates it rather than simplifying and Cesaro Stolz doesn't seem to work either.

copper.hat
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user313117
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    Might be related. http://math.stackexchange.com/questions/150391/evaluate-lim-limits-n-to-infty-frac-sum-k-1n-kmnm1 –  Feb 13 '16 at 09:21
  • @frank000 1. That might constitute a duplicate. 2. How did you find that so quickly? I need to learn the secret. (serious question) :) – Em. Feb 13 '16 at 09:29
  • Would the double limits be equal when replacing m by n? – user313117 Feb 13 '16 at 09:33
  • @probablyme I google $1^k+2^k+...+n^k$ and found it but as the answer suggest those questions are quite different actually. –  Feb 13 '16 at 09:36
  • @frank000 Yes, I see the difference. I just glanced. Did you google with plain text 1^k+2^k+...+n^k? – Em. Feb 13 '16 at 09:41
  • @probablyme Yes. –  Feb 13 '16 at 09:43
  • @frank000 Oooo, ok, thanks! – Em. Feb 13 '16 at 09:43
  • You can search on this site for https://math.stackexchange.com/search?q=%22\frac{1^n%2B2^n%2B%22+%22n^n}{n^n}%22 to find duplicates and directly related problems. – Lutz Lehmann Feb 13 '16 at 09:58
  • @LutzL: Nice find. – copper.hat Feb 13 '16 at 10:02
  • @LutzL: I have used up my supply of nice/awesome comments for the day :-). – copper.hat Feb 13 '16 at 10:15
  • @deleters : It seems you do not know your limits!! We do not need spoilers on this website!! – Mhenni Benghorbal Feb 15 '16 at 03:14
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    @MhenniBenghorbal: What you call your answer provides no additional information whatsoever (the OP has mentioned that they have already considered l'Hôpital). When asked for hints (by other high reputation users) as to how one can apply l'Hôpital in this case you side step by suggesting that others look at your other answers. I looked at over 100 of your answers, and searched, but found nothing dealing with l'Hôpital relevant to this problem. You are a high rep. user with a non answer and refuse to provide any more information elaborating your answer. What do you expect? – copper.hat Feb 15 '16 at 16:56
  • There is also this topic : http://math.stackexchange.com/questions/927771/what-is-lim-n-to-infty-sum-k-1n-left-frackn-rightn – JeSuis Feb 16 '16 at 18:51

2 Answers2

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Bernoulli's Inequality says that for $n\ge k$, $$ \left(1-\frac kn\right)^n $$ is an increasing sequence. Therefore, by Monotone Convergence $$ \begin{align} \sum_{k=0}^n\left(\frac kn\right)^n &=\sum_{k=0}^n\left(\frac{n-k}n\right)^n\\ &=\sum_{k=0}^n\left(1-\frac kn\right)^n\\ &\to\sum_{k=0}^\infty e^{-k}\\ &=\frac e{e-1} \end{align} $$

robjohn
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  • My question is this... What does Bernoulli's inequality and monotone convergence have to do with it? I mean, you could have started from that summation point. The LHS counts upwards and the RHS counts downwards. And from there you could write it as a GP of e. – Saikat Feb 13 '16 at 10:57
  • One needs to justify the interchange of limit and infinite sum (which is itself a limit) in $$\lim_{n\to\infty}\sum_{k=1}^\infty f_n(k) =\sum_{k=1}^\infty\lim_{n\to\infty}f_n(k)$$ Monotone Convergence does this. To show that the given sequence is Monotonic, we can use Bernoulli. – robjohn Feb 13 '16 at 11:05
  • Would the downvoter care to comment? – robjohn Feb 13 '16 at 13:27
  • Using the MCT is a a great idea here. – copper.hat Feb 13 '16 at 16:20
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$$\lim_{n \to \infty} \dfrac{ 1^n +2^n+\cdots +n^n}{n^n} = \lim_{n \to \infty}\frac{n^n}{n^n}+\frac{(n-1)^n}{n^n}+\frac{(n-2)^n}{n^n}+\cdots$$ $$=\lim_{n \to \infty}1+(1-1/n)^n+(1-2/n)^n +\cdots=1+e^{-1}+e^{-2}+\cdots$$ then one can sum the geometric series.

YC Chen
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    Simple and elegant +1 and accepted – user313117 Feb 13 '16 at 09:35
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    The error of $(1-k/n)^n-e^{-k}=e^{-k}(e^{k+n·\ln(1-k/n)}-1)=e^{-k}·(k^2/n+k·O((k/n)^2)$ does not seem negligible in a sum with growing number of summands. – Lutz Lehmann Feb 13 '16 at 09:35
  • I too found every bit of this solution elegant. Right from reversing the series to writing it as a GP of e. – Saikat Feb 13 '16 at 10:43
  • "Elegant", perhaps, in need of a serious justification for the interversion of limits and summation, no doubt! (This is merely repeating @LutzL's observation above, which, rather amazingly, was not even addressed in the least until now.) – Did Feb 14 '16 at 22:29