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

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

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

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.

• thank you for the explanation Feb 10, 2019 at 14:14

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

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

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

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

• +1. This avoid to deal with the factorial of a negative number. Feb 10, 2019 at 14:24

Hint:$$\dfrac{k^2}{k!}=\dfrac{k}{(k-1)!}=\dfrac{k-1}{(k-1)!}+\dfrac{1}{(k-1)!}.$$

• thank you, that's what I needed Feb 10, 2019 at 14:14
• @chendoytshman You welcome :) Feb 10, 2019 at 14:19

Hint:

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

• that's what I needed, thank you Feb 10, 2019 at 14:14

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