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?
 A: By definition of $e$, we have
$$
e = \sum_{n=0}^\infty \frac{1}{n!}
= \sum_{n=0}^\infty \frac{1}{(2n+1)!}+\sum_{n=0}^\infty \frac{1}{(2n)!}
$$
by separating even and odd terms (both series still converge).
Bringing them together again,
$$\begin{align}
e &= \sum_{n=0}^\infty \left( \frac{1}{(2n+1)!}+\frac{1}{(2n)!}\right)
= \sum_{n=0}^\infty \frac{(2n)!+(2n+1)!}{(2n+1)!(2n)!} \\
&= \sum_{n=0}^\infty \frac{1+(2n+1)}{(2n+1)!}
= \sum_{n=0}^\infty \frac{2(n+1)}{(2n+1)!}
\end{align}$$
giving you the result.
(Note that $\sum_{n=0}^\infty \frac{2(n+1)}{(2n+1)!}=\sum_{n=1}^\infty \frac{2n}{(2n-1)!}$ by a simple change of indices.)
A: Start from the known series for $e$
$$e=\sum_{j=0}^\infty \frac{1}{j!}$$
and group every even term with the next one :
$$e=\sum_{j=0}^\infty \frac{1}{j!}=\sum_{k=0}^\infty \frac{1}{(2k)!}+\frac 1{(2k+1)!}=\sum_{k=0}^\infty \frac{2k+1}{(2k+1)!}+\frac{1}{(2k+1)!}=\sum_{k=0}^\infty \frac{2k+2}{(2k+1)!}.$$
A: $$\frac{2n+2}{\left(2n+1\right)!}=\frac{2n+1}{(2n+1)!}+\frac{1}{(2n+1)!}=\frac{1}{(2n)!}+\frac{1}{(2n+1)!}$$
so
$$\sum _{n=0}^{\infty \:}\cfrac{2n+2}{\left(2n+1\right)!}=\sum _{n=0}^{\infty \:}\frac{1}{(2n)!}+\frac{1}{(2n+1)!}=\cosh 1+\sinh 1=e$$
A: Note that
$$\begin{align}
\frac 2{1!}&=\frac 1{0!}+\frac 1{1!}\\
\frac 4{3!}&=\frac 1{2!}+\frac 1{3!}\\
\frac 6{5!}&=\frac 1{4!}+\frac 1{5!}\\
\frac 8{7!}&=\frac 1{8!}+\frac 1{7!}\\
&\vdots\\
\frac {2n}{(2n-1)!}&=\frac 1{(2n-2)!}+\frac 1{(2n-1)!}
\end{align}$$
Summing for $n=1$ to $\infty$ for RHS gives $e$ which is a well-known result.
Hence the sum for LHS also equals $e$, which is the required proof. 

Alternatively 
$$\begin{align}
e=\sum_{k=0}^\infty \frac 1{k!}&=\sum_{n=0}^\infty \frac 1{(2n)!}+\frac 1{(2n+1)!}\\
&=\sum_{n=0}^\infty \frac 1{(2n)!}+\frac 1{(2n+1)(2n)!}\\
&=\sum_{n=0}^\infty \frac 1{(2n)!}\left(1+\frac 1{2n+1}\right)\\
&=\sum_{n=0}^\infty \frac {2n+2}{(2n+1)(2n)!}\\
&=\sum_{n=\color{red}0}^\infty \frac {2n\color{red}{+2}}{(2n\color{red}{+1})!}\\
&=\sum_{n=\color{red}1}^\infty \frac {2n}{(2n\color{red}{-1})!}\quad\blacksquare\\
\end{align}$$
A: 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, $S=e$.
