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I'm having trouble convincing myself why

$$\sum_{k = 0}^{\infty} \frac{k}{k!} = e.$$

As I was under the impression that only

$$\sum_{k = 0}^\infty \frac{1}{k!} = e$$

by definition.

By writing out terms of the first series

$$\sum_{k = 0}^{\infty} \frac{k}{k!} = 0 + 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + ...$$

Shouldn't it then be $1 + e$, not just $e$?

Thanks!

Featherball
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    Why? $0+1+1+\frac12+\frac16+\ldots=1+1+\frac12+\frac16+\ldots$ – Hagen von Eitzen Apr 15 '20 at 13:02
  • Note that the first two terms of $\sum_{k=0}^\infty \frac{1}{k!}$ are $\frac{1}{0!} = 1$ and $\frac{1}{1!} = 1$. – user771918 Apr 15 '20 at 13:07
  • https://math.stackexchange.com/questions/1711318/show-sum-k-1-infty-frack2k-2-mathrme https://math.stackexchange.com/questions/576976/evaluate-the-series-lim-limits-n-to-infty-sum-limits-i-1n-fracn22 – lab bhattacharjee Apr 15 '20 at 13:21

1 Answers1

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Since $\frac 0{0!}=0$, we have $$\sum_{k=0}^\infty\frac k{k!}= \sum_{k=1}^\infty\frac k{k!}=\sum_{k=0}^\infty\frac {k+1}{(k+1)!}=\sum_{k=0}^\infty\frac 1{k!}=e.$$

Hagen von Eitzen
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