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?

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