2
$\begingroup$

I have some trouble with series theory. The specific questions are as follows: \begin{equation} \sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!} \end{equation} 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.

$\endgroup$
2
  • 3
    $\begingroup$ how did you get from $(2n)!$ to $2^nn!$? $\endgroup$ Commented Aug 17, 2020 at 4:45
  • $\begingroup$ $(2n)!!=2^n n!$ Thank you, it's my fault. $\endgroup$ Commented Aug 17, 2020 at 4:52

2 Answers 2

3
$\begingroup$

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)!} . $

$\endgroup$
2
$\begingroup$

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .