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

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

A: Hint:$$\dfrac{k^2}{k!}=\dfrac{k}{(k-1)!}=\dfrac{k-1}{(k-1)!}+\dfrac{1}{(k-1)!}.$$
A: Hint:
Consider 
$$k^2=k(k-1)+k.$$
A: 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 
$$\sum_{k=0}^\infty \frac{k^2}{k!} = \sum_{k=1}^\infty \frac{k^2}{k!} = e^1(1+1) = 2e$$
