# How can I find the limit without using a closed form expression [duplicate]

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. TaussigFeb 13 at 13:46

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

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

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Simple and elegant +1 and accepted – user313117 Feb 13 at 9:35
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