# 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.

• Hint: if you know $e^x$ is 'all factorials', how could you get 'just the odd ones'? – Integrand Aug 7 at 0:51

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!

• Could you expand on step 2 of methodology 2. I do not find that easy to see the separation you did from the fraction involving $(-1)^n$ to two different fractions. – Samuel A. Morales Aug 7 at 13:06
• Sure. This was a decomposition of even and odd terms. But the odd terms are negative while the even terms are positive. – Mark Viola Aug 7 at 13:10
• Thank you @Mark Viola. It helped my a lot to solve my problem. Also thanks for your willingness to help. Already marked it has the answer! – Samuel A. Morales Aug 7 at 13:13

Hint: $$\frac{n}{(2n+1)!} = \frac{2n+1-1}{2(2n+1)!} = \frac{1}{2 (2n)!} - \frac{1}{2(2n+1)!}$$

• +1 That avoids any further answer. Nice job. – Felix Marin Aug 13 at 21:32
• @FelixMarin at the end of the day it's nice to know there's something you know – Alex Aug 14 at 10:59

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*}

\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*}
\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 $$\mathbb{R}$$.
• WA gives $S'(1)=\frac{\sqrt{e}}{2}$... – Alexey Burdin Aug 7 at 1:22
• This is not the correct result. For $x<0$, the $\sqrt{x}$ is not defined on the reals. And $e^{\sqrt z}$ has no Taylor series representation for $z\in \mathbb{C}$? If a function has a Taylor series, it is analytic. $e^{\sqrt z}$ is not analytic. – Mark Viola Aug 7 at 2:38