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how do I show that

$$\frac{1}{2}\cdot \sum _{k=0}^{\infty }\:\frac{k^2}{k!}\:=e$$

I don't know how to manage the product inside the sum in order to calculate the sigma. any hints? thanks

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  • $\begingroup$ Too many parentheses make reading cumbersome. I also changed your index $\;x\;$ to index $\;k\;$ . I think $\;x\;$ makes things a little more confusing. $\endgroup$
    – DonAntonio
    Feb 10, 2019 at 14:08
  • $\begingroup$ Is there anything which could be improved which prevents you from accepting one of the given answers? $\endgroup$
    – mrtaurho
    May 19, 2019 at 18:39
  • $\begingroup$ forgot to do so, just did. $\endgroup$
    – chendoy
    May 22, 2019 at 21:51

5 Answers 5

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Notice that:

$$\frac{0^2}{0!} = 0, \frac{1^2}{1!} = 1.$$

Then: $$\frac{1}{2} \sum _{k=0}^{\infty }\:\frac{k^2}{k!} = \frac{1}{2}\left(0 + 1 + \sum _{k=2}^{\infty }\frac{k^2}{k!}\right).$$

Therefore:

$$\frac{1}{2} \sum _{k=0}^{\infty }\:\frac{k^2}{k!} = \frac{1}{2}\left(1 + \sum _{k=2}^{\infty }\frac{1}{(k-1)!} + \sum _{k=2}^{\infty }\frac{1}{(k-2)!}\right) = \\ = \frac{1}{2}\left(1 + \sum _{j=1}^{\infty }\frac{1}{j!} + \sum _{h=0}^{\infty }\frac{1}{h!}\right),$$

where $j= k-1$ in the first sum, and $h = k-2$ in the second.

Finally: $$\frac{1}{2}\left(1 + \sum _{j=0}^{\infty }\frac{1}{j!} - 1 + \sum_{h=0}^{\infty }\frac{1}{h!}\right) = \\ = \frac{1}{2}\left(1 + e - 1 + e\right) = e.$$

As a final remark, notice that, in this way, we avoid to deal with the factorial of a negative integer.

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  • $\begingroup$ thank you for the explanation $\endgroup$
    – chendoy
    Feb 10, 2019 at 14:14
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$$e^x=\sum_{k=0}^\infty \frac{x^k}{k!}\implies \frac{\mathrm d}{\mathrm dx}e^x = \frac{\mathrm d}{\mathrm dx}\sum_{k=0}^\infty \frac{x^k}{k!}\implies e^x=\sum_{k=0}^\infty \frac{k}{k!}x^{k-1}$$

$$xe^x=\sum_{k=0}^\infty \frac{k}{k!}x^k\implies\frac{\mathrm d}{\mathrm dx}xe^x=\frac{\mathrm d}{\mathrm dx}\sum_{k=0}^\infty \frac{k}{k!}x^k\implies (x+1)e^x =\sum_{k=0}^\infty \frac{k^2}{k!}x^{k-1}$$

$$\therefore~(1+1)e^1=\sum_{k=0}^\infty \frac{k^2}{k!}1^{k-1}$$

$$\therefore~\frac12\sum_{k=0}^\infty \frac{k^2}{k!}=e$$

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    $\begingroup$ +1. This avoid to deal with the factorial of a negative number. $\endgroup$ Feb 10, 2019 at 14:24
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Hint:$$\dfrac{k^2}{k!}=\dfrac{k}{(k-1)!}=\dfrac{k-1}{(k-1)!}+\dfrac{1}{(k-1)!}.$$

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    $\begingroup$ thank you, that's what I needed $\endgroup$
    – chendoy
    Feb 10, 2019 at 14:14
  • $\begingroup$ @chendoytshman You welcome :) $\endgroup$ Feb 10, 2019 at 14:19
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Hint:

Consider $$k^2=k(k-1)+k.$$

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  • $\begingroup$ that's what I needed, thank you $\endgroup$
    – chendoy
    Feb 10, 2019 at 14:14
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We have $$xe^x = \sum_{k=0}^\infty \frac{x^{k+1}}{k!} = \sum_{k=1}^\infty \frac{x^k}{(k-1)!} = \sum_{k=1}^\infty \frac{kx^k}{k!}$$ so taking the derivative yields $$e^x(x+1) = \sum_{k=1}^\infty \frac{k^2x^{k-1}}{k!}$$

Plugging $x = 1$ finally gives $$\sum_{k=0}^\infty \frac{k^2}{k!} = \sum_{k=1}^\infty \frac{k^2}{k!} = e^1(1+1) = 2e$$

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