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I'm having trouble seeing why $\sum_{k=1} k^{2}/k!=e^2$. The ratio test says that it converges absolutely. Pardon my ignorance, but are there any techniques to show this?

I thought about expanding $e^{2}= e \cdot e =(\sum_{k=0} \frac{1}{k!})(\sum_{k=0} \frac{1}{k!})$ but so far that hasn't helped me much.

user135520
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4 Answers4

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You may have misread your source, this is not equal to $e^2$, but to $2e$. Unrolling the expression, $$\begin{align} \sum_{k=0}^\infty \frac{k^2}{k!} &= \sum_{k=1}^\infty \frac{k^2}{k!} = \sum_{k=1}^\infty \frac{k}{(k-1)!} = \sum_{k=0}^\infty \frac{k+1}{k!} = \sum_{k=0}^\infty \frac{k}{k!} +\sum_{k=0}^\infty \frac{1}{k!} \\ & = \sum_{k=1}^\infty \frac{k}{k!} +e^1 = \sum_{k=1}^\infty \frac{1}{(k-1)!} +e^1 = \sum_{k=0}^\infty \frac{1}{k!} +e^1 = e^1 +e^1 \\ &= 2e \end{align}$$

Clement C.
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\begin{align*} \sum_{k=0}^\infty \frac{k^2}{k!}&=\sum_{k=1}^\infty \frac{k}{(k-1)!} =\sum_{k=1}^\infty \frac{k-1}{(k-1)!}+\sum_{k=1}^\infty \frac{1}{(k-1)!} \\ &= \sum_{k=2}^\infty \frac{k-1}{(k-1)!}+\sum_{k=0}^\infty \frac{1}{k!}=\sum_{k=2}^\infty \frac{1}{(k-2)!}+e =\sum_{k=0}^\infty \frac{1}{k!}+e =2e \end{align*}

Yiorgos S. Smyrlis
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$ e^x= \sum \frac{x^k}{k!} $

$ e^x= (e^x)'= \sum \frac{x^{(k-1)}}{(k-1)!} $

$ x e^x = \sum \frac{x^k}{(k-1)!} $

$ (x e^x)' = (x+1)e^x = \sum \frac{k x^{(k-1)}}{(k-1)!} = \sum \frac{k^2 x^{(k-1)}}{k!} $

so $2e = \sum \frac{k^2}{k!}$

qwenty
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No. Actually,

$\sum_{k=0} 2^{k}/k!=e^2$

Will Jagy
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    The fact that another series has sum $e^2$ does not mean that this one cannot sum to $e^2$. (In this case, it does not, but the argument is not a valid one.) – Clement C. Sep 04 '16 at 23:04
  • @ClementC. sure. But many people throw random slop here without proofreading. I might have worried, but you and another found the actual value of the series as typed. – Will Jagy Sep 04 '16 at 23:10