6
$\begingroup$

\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}

$\endgroup$
  • $\begingroup$ 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 $n^n$? $\endgroup$ – CuriousGuest Oct 11 '14 at 17:26
10
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Thanks a lot!It's the right answer. $\endgroup$ – GEE20151011 Oct 11 '14 at 17:34
  • $\begingroup$ 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 $\endgroup$ – mvggz Oct 12 '14 at 10:24
  • $\begingroup$ @mvggz Thank you for correction! I edited the answer. $\endgroup$ – user2097 Oct 12 '14 at 14:01
  • $\begingroup$ 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) ? $\endgroup$ – mvggz Oct 12 '14 at 14:22
  • 1
    $\begingroup$ @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 $1+1/e+1/e^2+...=e/(e-1).$ $\endgroup$ – user2097 Oct 12 '14 at 14:41
0
$\begingroup$

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:

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.