Someone has posted a question about evaluating $$ A \colon =\lim_{\mathbb N^* \ni n \to \infty }\frac {1^n + 2^n + \cdots + n^n} {n^n}, $$ and I thought I could apply Cesaro-Stolz theorem, because the denominator $n^n \nearrow +\infty$. But if I apply it, then I get $$ A = \lim_n \frac {(n+1)^{n+1}} {(n+1)^{n+1} - n^n} = \lim_n \frac {\left( 1 + \dfrac 1n\right)^{n+1}} {\left( 1 + \dfrac 1n\right)^{n+1} - \dfrac 1n} = \frac {\mathrm e}{\mathrm e - 0} = 1, $$ instead of $\mathrm e / (\mathrm e - 1)$. What happened here?

Possible duplicate from following post: Understanding $(\frac{1}{n})^n+(\frac{2}{n})^n+...+(\frac{n}{n})^n$ sum

Older post: How to evaluate $ \lim \limits_{n\to \infty} \sum \limits_ {k=1}^n \frac{k^n}{n^n}$?

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    $\begingroup$ the numerator is not $(n+1)^{n+1}$. $\endgroup$ – Nosrati Aug 8 '18 at 4:42
  • $\begingroup$ @user108128 Thanks, I notice it now. $\endgroup$ – xbh Aug 8 '18 at 4:44
  • $\begingroup$ But why is the denominator not $(n+1)^{n+1}$? $\endgroup$ – Szeto Aug 9 '18 at 15:09
  • $\begingroup$ @Szeto It should be $$ \sum_1^{n+1} j^{n+1} - \sum_1^n j^n$$, which clearly does not equal $(n+1)^{n+1}$ $\endgroup$ – xbh Aug 9 '18 at 15:10
  • $\begingroup$ @xbh Ah! I even typed ‘denominator’ in my comment! I better get some rest. $\endgroup$ – Szeto Aug 9 '18 at 15:14

Thanks to @user108128. This is purely my mistake. Maybe this is a lesson for someone read the post could learn [or not, cause this mistake is the “inferior” type].

  • $\begingroup$ Please elaborate what @user108128 stated as soln., as his response is not any more visible. $\endgroup$ – jiten Sep 11 '18 at 10:14
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    $\begingroup$ Basically the same as Nosrati stated. The explanations were also given in the following comment. $\endgroup$ – xbh Sep 11 '18 at 10:17

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