# Tricky Infinite Sum

1. Dec 19, 2009

### stanli121

1. The problem statement, all variables and given/known data

What is $$\sum_{n=0}^{\infty}{\frac{1}{(2n)!}}$$?

2. Relevant equations
$$e^x$$ = $$\sum_{n=0}^{\infty}{\frac{x^n}{n!}}$$

3. The attempt at a solution
I understand that the given series looks like the series for e^1 but I know that isn't the correct answer. The answer (without solution) was supplied as (1/2)(e+e^-1). I can't seem to figure out the extra e^-1 part. Help?

2. Dec 19, 2009

### sylas

Try writing out e (which equals e1) and e-1 as infinite sums, using the equation you have given.

Cheers -- sylas

3. Dec 19, 2009

### lanedance

have you treid working backwards from the answer to understand how they get there?

4. Dec 20, 2009

### HallsofIvy

Perhaps more to the point
$$cos(x)= \sum_{n=0}^\infty \frac{(-1)^nx^n}{(2n)!}$$

What x gives $(-1)^nx^n= (-x)^n= 1$?

5. Dec 20, 2009

### sylas

Um, actually
$$cos(x)= \sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n)!}$$
This is not going to help. And let's not introduce the hyperbolic cos. He's got the answer, and now just needs to work back from there.

Cheers -- sylas