I am trying to solve the following limit, which includes a sum:

$\lim_{n\to\infty} [\frac{1}{n^2}(2+\frac{3^2}{2}+\cdots+\frac{(n+1)^n}{n^{n-1}})]$

So far, the only thing I have been able to do is to bound the limit between 0 and e:

  • If $(2+\frac{3^2}{2}+\cdots+\frac{(n+1)^n}{n^{n-1}}) = 0 \to \lim_{n\to\infty} \frac{1}{n^2}=0$
  • If $(\frac{(n+1)^n}{n^{n-1}}+\cdots+\frac{(n+1)^n}{n^{n-1}}) = n \frac{(n+1)^n}{n^{n-1}} \to \lim_{n\to\infty} \frac{n}{n^2}\frac{(n+1)^n}{n^{n-1}}= \lim_{n\to\infty} \frac{(n+1)^n}{n^n} = \lim_{n\to\infty} (1+\frac{1}{n})^n = e $

$0 < (2+\frac{3^2}{2}+\cdots+\frac{(n+1)^n}{n^{n-1}}) < (\frac{(n+1)^n}{n^{n-1}}+\cdots+\frac{(n+1)^n}{n^{n-1}}) \to\\ 0 < \lim_{n\to\infty} [\frac{1}{n^2}(2+\frac{3^2}{2}+\cdots+\frac{(n+1)^n}{n^{n-1}})] < e$

This is fine, but I would like to know if a closer bound or an exact limit could be found.

Thanks in advance!

  • $\begingroup$ I'm not exactly sure how they found this answer, but I typed the limit into Wolfram Alpha, and they said the answer was $0$. $\endgroup$ – Noble Mushtak Oct 25 '16 at 23:00
  • $\begingroup$ where did the factor $\frac{1}{n^2}$ go when you calculated the bounds? $\endgroup$ – user159517 Oct 25 '16 at 23:16
  • $\begingroup$ The factor is used when calculating the limits itself. $\endgroup$ – Alex Martínez Ascensión Oct 26 '16 at 5:21

The expression equals

$$\frac{\sum_{k=1}^{n} (k+1)^k/k^{k-1}}{n^2}.$$

Stolz-Cesaro says we should look at

$$\frac{(n+2)^{n+1}/(n+1)^n}{(n+1)^2 - n^2}=\frac{(1+1/(n+1))^n(n+2)}{2n+2}.$$

The last expresion $\to \frac{e}{2}.$ So by S-C, the limit is $e/2.$


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