Evaluating the sum $\sum\limits_{n=1}^{\infty}\frac{x^{2n}}{(2n+1)!}$

Question

I'm having difficulties with the sum above. My first attempt was to rewrite it like $$ \sum\limits_{n=1}^{\infty}\frac{x^{2n}}{(2n+1)!}=\frac{1}{x}\sum\limits_{n=1}^{\infty}\frac{x^{2n+1}}{(2n+1)!} $$ and hence use substitution $k=2n+1$, so I got $$ \frac{1}{x}\sum\limits_{k=3}^{\infty}\frac{x^k}{k!}=\frac{1}{x}\left(\sum\limits_{k=0}^{\infty}\frac{x^k}{k!}-1-x-\frac{x^2}{2}\right)=\frac{1}{x}\left(e^x-1-x-\frac{x^2}{2}\right), $$ as I was trying to use series for exponential. The thing is that this is a way different result from the result which I would get if I've used series for $\sinh(x)$ (my intention was to avoid it). It's still not clear to me how this is possible. Is it because of the substitution? Could anybody explain please? The result I should get is $$ \frac{1}{2x}e^x-\frac{1}{2x}e^{-x}-1. $$

Answer 1

It looks like some calculation error. If you consider the sum from $n=0$ to $\infty$ then it equals $$\frac{1}{x}( \frac{e^x - e^{-x}}{2})$$

If the sums now runs from $1$ to $\infty$ then one term $=1$ is missing, hence you get the result.

Answer 2

Yes, that is because of the substitution. whenu put k=2n+1, you get n=$\frac{k-1}{2}$ and for n to be integer k must be odd so it doesn't take all the values from k=3 to $\infty$ but only the odd values and thus you must subtract the two exponential series to get them

Answer 3

The well known series expansion for $\sinh x$ is

$$\sinh x = \sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)!},$$

Dividing by $x$:

$$\frac{\sinh x }{x}= \sum_{n=0}^\infty \frac{x^{2n}}{(2n+1)!},$$

so your series is

$$\frac{\sinh x}{x}-1 = \sum_{n=1}^\infty \frac{x^{2n}}{(2n+1)!}.$$

Answer 4

Let $\sum_{n=1}^\infty a_n$ be absolutely convergent. Then, we have the identity

$$\sum_{n=1}^\infty a_{2n-1}=\sum_{n=1}^\infty \left(\frac{1-(-1)^n}{2}\right)a_n\tag1$$

Next, let $a_n=\frac{x^{n+2}}{(n+2)!}$ in $(1)$ to find

$$\begin{align} \sum_{n=1}^\infty \frac{x^{2n}}{(2n+1)!}&=\sum_{n=1}^{\infty}\left(\frac{1-(-1)^n}{2}\right)\frac{x^{n+1}}{(n+2)!}\\\\ &=\frac12\sum_{n=1}^{\infty}\frac{x^{n+1}}{(n+2)!}+\frac12 \sum_{n=1}^\infty \frac{(-x)^{n+1}}{(n+2)!}\\\\ &=\frac1{2x}\sum_{n=1}^{\infty}\frac{x^{n+2}}{(n+2)!}-\frac1{2x}\sum_{n=1}^{\infty}\frac{(-x)^{n+2}}{(n+2)!}\\\\ &=\frac1{2x}\left(e^x-1-x-\frac12x^2\right)-\frac1{2x}\left(e^{-x}-1+x-\frac12x^2\right)\\\\ &=\frac{\sinh(x)}{x}-1 \end{align}$$