I'm not sure which section a post on finding the Euler numbers using series belongs, so I put it here.

Find (a) the radius of convergence, and (b) the first six Euler numbers $\displaystyle E_n$ from:

$\displaystyle \frac1{\cosh(z)}=\sum_{n=0}^\infty \frac{E_n}{n!}z^n$

I haven't any idea how to start the part (a).

I can't put the finishing touches on part (b):

$\displaystyle \cosh(z) = \frac{e^z+e^{-z}}{2}$

$\displaystyle e^z = \sum_{n=0}^\infty \frac{z^n}{n!}$

$\displaystyle e^{-z} = \sum_{n=0}^\infty \frac{(-1)^n z^n}{n!}$

$\displaystyle e^z+e^{-z} = \sum_{n=0}^\infty \frac{z^n}{n!} + \sum_{n=0}^\infty \frac{(-1)^n z^n}{n!} = \sum_{n=0}^\infty \frac{(2z)^n}{(2n)!}$

So,

$\displaystyle \frac{1}{\cosh(z)} = \frac{2}{2\sum_{n=0}^\infty \frac{(2z)^n}{(2n)!}} = \frac{1}{\sum_{n=0}^\infty \frac{(2z)^n}{(2n)!}}$

However, I can't quite get this to a format where I can get the Euler numbers.

I know I can't flip the bottom fraction since it's an infinite sum. But I only need the first six, yet I still can't do it. The first one I can do with the first term of the series (right?):

$\displaystyle \frac{1}{\cosh(z)} = \frac{1}{\sum_{n=0}^\infty \frac{z^n}{(2n)!}} = \frac{1}{1}=1$ and $\displaystyle E_0=1$ (so the back of my book says).

The next term, and the one after that, is:

$\displaystyle \frac{1}{\cosh(z)} = \frac{1}{1+z^2/2}=\frac{2}{2+z^2}$

$\displaystyle \frac{1}{\cosh(z)} = \frac{1}{1+z^2/2+z^4/4!}=\frac{24}{24+12z^2+z^4}$

Now what? How do I find the coefficient's of z, z^2, etc. from this?

I'm still working on part (a), so I don't have a specific question yet.

Thanks.

P.S. How do you make a new line in this forum? I tried \\, \\*, and \newline with no success and had to make separate 'math' tags for each line.