How to compute $\sum_{n=1}^\infty{\frac{n}{(2n+1)!}}$? In a calculus book I am reading I have encountered the following problem:
$$\sum_{n=1}^\infty{\frac{n}{(2n+1)!}}$$
The hint is to use Taylor series expansion's for $e^x$. I tried to express the sum as the form $$e^x=\sum_{n=0}^{\infty}{\frac{x^n}{n!}}$$
But I could not find a consistent method, I always end un with different sums of factorials that does not help me solve the problem
The official solution is $$\boxed{\frac{1}{2e}}$$
The excersise is in a chapter that mixes calculus with summation, so the solution will probably include both.
Any help or hint is highly appreaciated! Thanks in advance.
 A: Hint: $\frac{n}{(2n+1)!} = \frac{2n+1-1}{2(2n+1)!} = \frac{1}{2 (2n)!} - \frac{1}{2(2n+1)!}$
A: METHODOLOGY $1$:
The Taylor series for the hyperbolic sine function $\sinh(x)$ is given by
$$\sinh(x)=\sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!}\tag1$$
If we divide both sides of $(1)$ by $x$, differentiate, and set $x=1$, awe find that
$$\underbrace{\cosh(1)-\sinh(1)}_{=e^{-1}}=2\sum_{n=1}^\infty\frac{n}{(2n+1)!}\tag2$$
Finally, dividing $(2)$ by $2$ yields the coveted result
$$\sum_{n=1}^\infty \frac{n}{(2n+1)!}=\frac1{2e}$$
as was to be shown!

METHODOLOGY $2$:
We begin with the Taylor series for $e^x$ at $x=-1$.  Then, we see that
$$\begin{align}
\frac1e&=\sum_{n=0}^\infty \frac{(-1)^n}{n!}\\\\
&=\sum_{n=0}^\infty\left(\frac1{(2n)!}-\frac1{(2n+1)!}\right)\\\\
&=\sum_{n=0}^\infty \frac{(2n+1)!-(2n)!}{(2n)!(2n+1)!}\\\\
&=\sum_{n=1}^\infty \frac{2n}{(2n+1)!}\\\\
&=2\sum_{n=1}^\infty \frac{n}{(2n+1)!}
\end{align}$$
from which we arrive at the coveted result
$$\sum_{n=1}^\infty \frac{n}{(2n+1)!}=\frac1{2e}$$
as expected!
A: First off, this looks like the derivative of the following sum, evaluated at $x = 1$:
$\begin{align*}
   &\sum_{n \ge 1} \frac{x^n}{(2 n + 1)!}
\end{align*}$
Trouble is that we get only odd terms of something like:
$\begin{align*}
   e^{x^{1/2}}
     &= \sum_{n \ge 0} \frac{x^{n/2}}{n!}
\end{align*}$
Now if:
$\begin{align*}
   f(z)
     &= \sum_{n \ge 0} a_n z^n
\end{align*}$
then:
$\begin{align*}
   \frac{f(z) + f(-z)}{2}
      &= \sum_{n \ge 0} a_{2 n} z^{2 n} \\
   \frac{f(z) - f(-z)}{2}
      &= \sum_{n \ge 0} a_{2 n + 1} z^{2 n + 1}
\end{align*}$
So your sum is:
$\begin{align*}
  S(x)
     &= x^{1/2} \frac{e^{x/2} - e^{-x/2}}{2} \\
     &= \sum_{n \ge 0} \frac{x^n}{(2 n + 1)!}
\end{align*}$
What is left is routine:
$\begin{align*}
   S'(1)
     &= \frac{e}{2} 
\end{align*}$
The manipulations are valid inside the region of convergence of the series for $e^x$, i.e., all of $\mathbb{R}$.
