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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.

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marked as duplicate by LutzL, Nehorai, Najib Idrissi, heropup, N. F. Taussig Feb 13 at 13:46

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Might be related. math.stackexchange.com/questions/150391/… – frank000 Feb 13 at 9:21
    
@frank000 1. That might constitute a duplicate. 2. How did you find that so quickly? I need to learn the secret. (serious question) :) – probablyme Feb 13 at 9:29
    
Would the double limits be equal when replacing m by n? – user313117 Feb 13 at 9:33
    
@probablyme I google $1^k+2^k+...+n^k$ and found it but as the answer suggest those questions are quite different actually. – frank000 Feb 13 at 9:36
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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 at 16:56
up vote 9 down vote accepted

$$\lim_{n \to \infty} \dfrac{ 1^n +2^n+\cdots +n^n}{n^n} = \lim_{n \to \infty}\frac{n^n}{n^n}+\frac{(n-1)^n}{n^n}+\frac{(n-2)^n}{n^n}+\cdots$$ $$=\lim_{n \to \infty}1+(1-1/n)^n+(1-2/n)^n +\cdots=1+e^{-1}+e^{-2}+\cdots$$ then one can sum the geometric series.

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Simple and elegant +1 and accepted – user313117 Feb 13 at 9:35
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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. – LutzL Feb 13 at 9:35
    
I too found every bit of this solution elegant. Right from reversing the series to writing it as a GP of e. – user230452 Feb 13 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 at 22:29

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

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+1: Nice. With justification too! – copper.hat Feb 13 at 9:42
    
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. – user230452 Feb 13 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 at 11:05
    
Would the downvoter care to comment? – robjohn Feb 13 at 13:27
    
Using the MCT is a a great idea here. – copper.hat Feb 13 at 16:20

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