In a calculus book I am reading I have encountered the following problem:


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.

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


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


Finally, dividing $(2)$ by $2$ yields the coveted result

$$\sum_{n=1}^\infty \frac{n}{(2n+1)!}=\frac1{2e}$$

as was to be shown!


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!

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  • $\begingroup$ 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. $\endgroup$ – Samuel A. Morales Aug 7 at 13:06
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    $\begingroup$ Sure. This was a decomposition of even and odd terms. But the odd terms are negative while the even terms are positive. $\endgroup$ – Mark Viola Aug 7 at 13:10
  • $\begingroup$ 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! $\endgroup$ – Samuel A. Morales Aug 7 at 13:13

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

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    $\begingroup$ +1 That avoids any further answer. Nice job. $\endgroup$ – Felix Marin Aug 13 at 21:32
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    $\begingroup$ @FelixMarin at the end of the day it's nice to know there's something you know $\endgroup$ – 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:

$\begin{align*} f(z) &= \sum_{n \ge 0} a_n z^n \end{align*}$


$\begin{align*} \frac{f(z) + f(-z)}{2} &= \sum_{n \ge 0} a_{2 n} z^{2 n} \\ \frac{f(z) - f(-z)}{2} &= \sum_{n \ge 0} a_{2 n + 1} z^{2 n + 1} \end{align*}$

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 $\mathbb{R}$.

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    $\begingroup$ WA gives $S'(1)=\frac{\sqrt{e}}{2}$... $\endgroup$ – Alexey Burdin Aug 7 at 1:22
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    $\begingroup$ 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. $\endgroup$ – Mark Viola Aug 7 at 2:38
  • $\begingroup$ I think your answer does not match the one given by the book. The method proposed by Mark Viola fits better the level I am expected. But thank you for you help and time! $\endgroup$ – Samuel A. Morales Aug 7 at 13:08

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