Show that there are complex numbers $E_2,E_4,E_6,\dotsc$ such that $\sec z = \frac{1}{\cos z} = 1+ \sum\limits_{k=1}^{\infty} \frac{E_{2k}}{(2k)!}z^{2k}$ in a neighborhood of $0$.
What is the radius of convergence?
Show that: $E_{2n}-{2n\choose 2n-2} E_{2n-2} + {2n\choose 2n-4}E_{2n-4}+ \dotsb -(-1)^{n}{2n \choose 2}E_{2} + (-1)^{n} = 0$.
Compute $E_2, E_4$ and $E_6$.
Can you show me how to solve this problem? I am not able to do any of them. Thanks.
Edit: Ideas have been given for 1,2,4. Thank you. What about item 3 ?