Solving Tricky Infinite Sum: What is \sum_{n=0}^{\infty}{\frac{1}{(2n)!}}?

[stanli121]

Homework Statement

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

Homework Equations

e^x = \sum_{n=0}^{\infty}{\frac{x^n}{n!}}

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?

 

[sylas]

Homework Statement

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

Homework Equations

e^x = \sum_{n=0}^{\infty}{\frac{x^n}{n!}}

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?

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

Cheers -- sylas

 

[lanedance]

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

 

[HallsofIvy]

Homework Statement

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

Homework Equations

e^x = \sum_{n=0}^{\infty}{\frac{x^n}{n!}}

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?

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?

 

[sylas]

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?

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