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# Find convergence series bellow:

Find convergence series above

Note by Pebrudal Zanu
3 years, 4 months ago

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The limit is $$\frac{e}{e-1}$$, the key is : $\sum_{k=1}^n \frac{k^n}{n} = \sum_{k=1}^n \left(1-\frac{k}{n} \right)^n$ Then you can do an asymptotic expansion for the summand and get the limit. · 2 years, 8 months ago