# Find the sum of series: $\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}$

I have some trouble with series theory. The specific questions are as follows: $$$$\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}$$$$ My idea is just like this:

Since $$e^x=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}$$, \begin{align} \sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}&=\sum_{n=0}^{\infty}\frac{x^{2n}}{2^nn!}\\ &=\sum_{n=0}^{\infty}\frac{(\frac{x^2}{2})^n}{n!}\\ &=e^{\frac{x^2}{2}} \end{align} However, the answer is cosh $$x$$. The main idea is based on the power series of $$e^x$$ and $$e^{–x}$$. Then add them together. But I still don't understand what I did wrong.

Can anyone help me out,please. Thank you.

• how did you get from $(2n)!$ to $2^nn!$? Aug 17, 2020 at 4:45
• $(2n)!!=2^n n!$ Thank you, it's my fault. Aug 17, 2020 at 4:52

What you did wrong was changing $$(2n)!$$ to $$2^nn!$$.

You were correct that $$e^x=\sum\limits_{n=0}^{\infty}\dfrac{x^{n}}{n!}$$,

so $$\cosh x = \dfrac{e^x+e^{-x}}2=\dfrac{\sum\limits_{n=0}^{\infty}\frac{x^{n}}{n!}+\sum\limits_{n=0}^{\infty}\frac{(-x)^{n}}{n!}}2=\dfrac{\sum\limits_{n=0}^{\infty}\frac{x^{n}}{n!}\left(1+(-1)^n\right) }2$$.

$$\dfrac{1+(-1)^n}2$$ is $$0$$ when $$n$$ is odd and $$1$$ when $$n$$ is even, so this becomes $$\sum\limits_{n=0}^{\infty}\dfrac{x^{2n}}{(2n)!} .$$

$$\sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}$$ $$=\frac{1}{2} \sum_{n=0}^{\infty} \frac{2 \cdot x^{2n}}{(2n)!} -\frac{1}{2} \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!} + \frac{1}{2} \sum_{n=0}^{\infty} \frac{ x^{2n+1}}{(2n+1)!}$$ $$=\frac{1}{2} \sum_{n=0}^{\infty} \frac{x^{n}}{(n)!} + \frac{1}{2} \sum_{n=0}^{\infty} \frac{(-x)^{n}}{(n)!}$$ $$= \frac{e^{x}+e^{-x}}{2}$$ $$= cosh(x)$$