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I want to compute $$l = \lim_{n\to\infty} \frac{1}{n^2} \left(\frac{2}{1} + \frac{9}{4} +\ ... \ +\frac{(n+1)^n}{n^{n-1}}\right)$$ My first thought was to use Riemann's sums, that is $\lim \frac{1}{n} \sum_{k=1}^n f(k/n) = \int_0^1 f(x)\ dx$. In order to apply this identity, I did this

$$l = \lim_{n\to\infty} \frac{1}{n^2} \sum_{k=1}^n \frac{(k+1)^k}{k^{k-1}} = \frac{1}{n^2} \sum_{k=1}^n k\frac{(k+1)^k}{k^{k}} = \frac{1}{n} \sum_{k=1}^n \frac{k}{n}\left( 1 + \frac{1}{k}\right)^k $$ But from this, I don't know how to follow. Any help is welcome.

Also I want to say that maybe this idea doesn't work to compute the limit. In that case, I will appreciate any hint to calculate $l$.

Thanks in advance.

Senna
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    The accepted answers in https://math.stackexchange.com/questions/3025640/evaluate-the-limit-of-lim-n-to-infty-frac1n2-left-frac21-frac9, https://math.stackexchange.com/questions/3248644/can-anyone-help-me-solve-these-sequence-limits, https://math.stackexchange.com/questions/1985151/find-the-following-limit-with-a-sum, https://math.stackexchange.com/questions/1607951/calculate-lim-n-to-infty-frac-frac21-frac322-frac4332 all suggest using Stolz–Cesàro. – player3236 Oct 29 '20 at 16:41

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