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\begin{aligned} \lim_{n \rightarrow \infty} {1^n+ 2^n +\cdots +n^n \over n^n} \end{aligned}

I tried to use the O'Stolz Rule.But it didn't work.

I was thinking about Riemann sums ,but I don't kown how to use it :(

Ps: \begin{aligned} a_1&=1^1\\ a_2&=1^2+2^2\\ &\vdots\\ a_n&=1^n+2^n+\cdots+n^n\\ a_{n+1}&=1^{n+1}+2^{n+1}+\cdots+n^{n+1}+(n+1)^{n+1} \end{aligned}

Michael Hardy
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GEE20151011
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  • After applying Stolz you get $\lim\frac{(n+1)^{n+1}}{(n+1)^{n+1}-n^n}$, have you tried dividing numerator and denominator by $n^n$? – CuriousGuest Oct 11 '14 at 17:26

2 Answers2

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Since $(1-k/n)^n<e^{-k}$, the limit equals $1+ \dfrac 1 e + \dfrac 1 {e^2} +\cdots=\dfrac e {e-1}$.

Michael Hardy
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user2097
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  • Thanks a lot!It's the right answer. – GEE20151011 Oct 11 '14 at 17:34
  • for k<n : 1-k/n <1, the term on the left of the inequality is bigger than the one on the right, so the inequality is false. And since the left part of it is true, you can't bound exp(-k) with your suit, so the proof doesn't hold – mvggz Oct 12 '14 at 10:24
  • @mvggz Thank you for correction! I edited the answer. – user2097 Oct 12 '14 at 14:01
  • Your welcome :) , but I still don't see how you can affirm that the limits of both series are the same. Don't you have just : L=< e/(e-1) ? – mvggz Oct 12 '14 at 14:22
  • @mvggz Exactly, my answer proves that $L\leqslant e/(e-1)$. Note that $\lim_{n\rightarrow\infty}(1-k/n)^n$ equals $e^{-k}$, so considering the $t$ greatest terms in nominator we get $L\geqslant 1+1/e+...+1/e^t$ for any $t$. – user2097 Oct 12 '14 at 14:32
  • I think there is something missing here, you can't affirm that L >=1+1/e+...+1/e^t for any t. The limit is made on n, for a given k but you can't say that if : U(n,k)--> V(k), when n--> ∞, and the series V(k) converges then the partial sum of U(n,k) does as well and both partial sums have the same limit, there is an argument missing, or at least something has to be added up. You know, it's like dominated convergence theorem – mvggz Oct 12 '14 at 14:38
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    @mvggz I just mean that, for any $t$, we have $1+1/e+...+1/e^t=\lim_{n \rightarrow \infty} {(n-t)^n+\cdots +n^n \over n^n}\leqslant \lim_{n \rightarrow \infty} {1^n+ 2^n +\cdots +n^n \over n^n}\leqslant1+1/e+1/e^2+...$, so the limit is $1+1/e+1/e^2+...=e/(e-1).$ – user2097 Oct 12 '14 at 14:41
  • Oh ok, my bad. Nice way to do it by the way, I had persuaded my mind that it was more sophisticated :) – mvggz Oct 12 '14 at 14:48
  • I think that's a bug, I can't edit my answer. Use the Cesaro-Stolz theorem to prove that this answer is not a solution.(the limit is $1$) – shooting-squirrel Oct 20 '14 at 03:13
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What you can do is a change of variable in the sum : $k'=n-k$

if you define:$ U(n,k) = (1-{k \over n})^n $and $V_n $ your sum, you have :

$$V_n = U(n,1) + ...+ U(n,n)$$

Now I would introduce this:$ E(x)$ is the floor of $x$

$$ U(n,k) = \int_k^{k+1} (1-{E(x) \over n})^n dx$$

So $$V_n= \int_1^n (1-{E(x) \over n})^ndx$$ Let $ T_n$ be: $(1-{E(x) \over n})^n$ if $ x=<n$, $ 0 $ if $x>n$

$$\Rightarrow V_n= \int_1^{+\infty} Tn(x) dx$$

$T_n(x)$ $\rightarrow$ $e^{-E(x)}$ when $n\rightarrow ∞$

You use the theorem that allows you to take the limit under the integrale, and you get:

$$\lim V_n = \int_1^{+\infty} e^{-E(x)} dx = {e \over e-1} $$

mvggz
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