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Now I want to find the limit $$\lim_{n\to\infty}\frac{\sum\limits_{k=1}^n k^n}{n^n}.$$ I try to use the Stolz theorem as follows: $$\lim_{n\to \infty}\frac{\sum_{k=1}^n k^n}{n^n}= \lim_{n\to \infty}\frac{\sum_{k=1}^{n+1} k^{n+1}-\sum_{k=1}^n k^n}{(n+1)^{n+1}-n^n}$$ $$=\lim_{n\to\infty}\frac{(n+1)^{n+1}+\sum_{k=1}^n(k^{n+1}-k^n)}{(n+1)^{n+1}-n^n}$$ $$=1+\lim_{n\to\infty}\frac{\sum_{k=1}^n(k^{n+1}-k^n)}{(n+1)^{n+1}}.$$ It seems to deal with the summation like this form: $$\sum_{k=1}^{n}k^p,\text{with}\ p=n,n+1.$$ I have no way to deal this summation, any help and hint will welcome!

Riemann
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2 Answers2

11

Hint: For fixed $k,$ $$\frac{(n-k)^n}{n^n}=\left(1-\frac{k}n\right)^n\to e^{-k}$$

Details:

Note that for $k=0...,n-1$ $$\log\left(1-\frac{k}{n}\right)=-\frac{k}{n}-\frac{k^2}{2n^2}-\cdots\leq -\frac{k}{n}$$

So $$\left(1-\frac{k}{n}\right)^n \leq e^{-k}$$

Then apply the dominated convergence theorem by defining:

$$f_n(k)=\begin{cases}0 & k\geq n\\ \left(1-\frac{k}{n}\right)^n& 0\leq k<n\end{cases}$$

Then $|f_n(k)|\leq g(k)=e^{-k}$ and for any $k,$ $\lim_{k\to\infty}f_n(k)\to g(k).$

Thomas Andrews
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10

We have that

$$\frac{\sum\limits_{k=1}^n k^n}{n^n}=\sum_{k=1}^n \left( \frac k n \right)^n=\sum_{k=0}^{n-1} \left( \frac {n-k} n \right)^n=\sum_{k=0}^{n-1} \left( 1-\frac {k} n \right)^n$$

then, following the suggestion given by ComplexYetTrivial, let consider

$$a_n=\sum_{\substack{k=0 \\ k<n}}^{\infty} \left( 1-\frac {k} n \right)^n$$

which is strictly increasing and bounded indeed by AM-GM we have

  • $\sqrt[n+1]{\left(1-\frac{k}{n}\right)^n \cdot 1} \leq \frac{n \left(1 - \frac{k}{n}\right) + 1}{n+1} = 1 - \frac{k}{n+1} $

and

  • $\left( 1-\frac {k} n \right)^n=e^{n\log \left( 1-\frac {k} n \right)}=e^{n \left( -\frac {k} {n}-\frac {k^2} {2n^2}-\frac {k^3} {3n^3}-\ldots \right)}\le e^{-k}$

therefore by the monotone convergence theorem $a_n$ has finite limit and since for $k$ fixed

  • $\left( 1-\frac {k} n \right)^n\to e^{-k}$

we have

$$\lim_{n\to \infty}\frac{\sum\limits_{k=1}^n k^n}{n^n}=\lim_{n\to \infty} \sum_{k=0}^{n-1} \left( 1-\frac {k} n \right)^n=\lim_{n\to \infty} \sum_{\substack{k=0 \\ k<n}}^{\infty} \left( 1-\frac {k} n \right)^n=\sum_{k=0}^{\infty}e^{-k}=\frac{e}{e-1}$$

user
  • 154,566
  • (+1) without Hints at last. :P – Mariusz Iwaniuk Jul 31 '18 at 14:56
  • @MariuszIwaniuk Thanks! The hint was already given! :) – user Jul 31 '18 at 15:01
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    In your second line, the $O(\frac{1}{n})$ depends on $k$. Therefore you can’t factorize $O(\frac{1}{n})$ in the sum without an additional argument. – mathcounterexamples.net Jul 31 '18 at 15:41
  • @mathcounterexamples.net Yes I think you are right, it's a step I need to revise. Thanks – user Jul 31 '18 at 15:44
  • You could write the sum as $\sum_{k=0}^\infty \left(1-\frac{k}{n}\right)^n \operatorname{1}(k<n)$, observe that the sequence $\left[\left(1-\frac{k}{n}\right)^n \operatorname{1}(k<n)\right]_{n\in\mathbb{N}}$ is increasing and use the monotone convergence theorem to interchange summation and limit. – ComplexYetTrivial Jul 31 '18 at 16:16
  • @ThomasAndrews By the AM-GM inequality we have $$ \sqrt[n+1]{\left(1-\frac{k}{n}\right)^n \cdot 1} \leq \frac{n \left(1 - \frac{k}{n}\right) + 1}{n+1} = 1 - \frac{k}{n+1} , . $$ – ComplexYetTrivial Jul 31 '18 at 16:28
  • Thanks for the discussion and suggestions, later I’ll to fix that part. Bye – user Jul 31 '18 at 17:02
  • @gimusi The estimates look fine, but I do not understand how you get $$\sum \limits_{k=0}^{n-1} \mathrm{e}^{-k \left(1-\frac{k}{n}\right)^{-1}} = \sum \limits_{k=1}^n \frac{n}{k} \mathrm{e}^{k-n} , . $$ – ComplexYetTrivial Aug 01 '18 at 01:52
  • @ComplexYetTrivial My approach was not conclusive, I've revised the proof following your suggestion. Thanks – user Aug 02 '18 at 06:02