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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. – Haroun Meghaichi · 2 years, 8 months ago

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TopNewestThe 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. – Haroun Meghaichi · 2 years, 8 months ago

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