<s>[INST] Prove that $ 2/1!+4/3!+6/5!+8/7!+\dots = e$ 
Prove that $ 2/1!+4/3!+6/5!+8/7!+\dots = e$

My attempt:
I googled the problem, and I found that $\sum_{(n=1)}^∞ \cfrac{2 n}{(2 n-1)!} = e$
I also found that $\sum _{n=0}^{\infty \:}\cfrac{2n+2}{\left(2n+1\right)!}$ is equal to $e$. 
How can I prove this?
 [/INST] Let $f(x)= x \sin(x)$. Then
$$
f(x) = \frac{x^2}{1!} - \frac{x^4}{3!} + \frac{x^6}{5!} -\frac{x^8}{7!} + \cdots
$$
and so
$$
f'(x) = \frac{2x}{1!} - \frac{4x^3}{3!} + \frac{6x^5}{5!} -\frac{8x^7}{7!} + \cdots
$$
Therefore
$$
f'(i) = \frac{2i}{1!} + \frac{4i}{3!} + \frac{6i}{5!} +\frac{8i}{7!} + \cdots = iS
$$
On the other hand,
$$
f'(x) = \sin(x) + x \cos(x)
$$
and so
$$
f'(i) = \sin(i) + i \cos(i) = i (\cos(-i) + i \sin(-i)) = ie^{i(-i)} = ie
$$
Therefore, $S=e$.
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