Taylor series of $\frac{2e^x}{e^{2x}+1}$ Consider the Taylor series of function $$\frac{2e^x}{e^{2x}+1}=\sum_{n=0}^\infty \frac{E_n}{n!}x^n$$
Prove that 
$$E_0=1, E_{2n-1}=0, E_{2n}=-\sum_{l=0}^{n-1}\binom{2n}{2l}E_{2l}, n\geq 1$$.
I am thinking to write the LHS as $\frac{1}{\frac{e^x+e^{-x}}{2}}=\frac{1}{\cosh x}$, but I am stuck here. How can I derive the above recurrence relation? Any enlightenment please?
 A: You want
$$
\left(\sum_{n=0}^\infty\frac{E_n}{n!}x^n\right)\cosh x=1
$$
i.e.
$$
\left(\sum_{n=0}^\infty\frac{E_n}{n!}x^n\right)
\left(\sum_{m=0}^\infty\frac1{(2m)!}x^{2m}\right)=1.
$$
Now expand the series in the LHS.  Clearly all $E_n$ for $n$ odd must vanish, or there will be a least $n$, $n$ odd, such that $E_n\neq 0$.  Then the coefficient of $x^n$ would not vanish.
So $E_0=1$, $E_{2n+1}=0$ for all $n$, and the coefficient of $x^{2n}$ gives
$$
\sum_{m=0}^n\frac{E_{2(n-m)}}{[2(n-m)]!(2m)!}=0
$$
which rearrange to the recurrence equation (remember $\frac{(2n)!}{[2(n-m)]!(2m)!}=\binom{2n}{2(n-m)}$).
A: note that:
$$\arctan(y)=\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}x^{2n+1}$$
and:
$$\frac{2e^x}{e^{2x}+1}=\frac{d}{dx}\left[\arctan(e^x)\right]$$
so:
$$\arctan(e^x)=\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}e^{(2n+1)x}$$
so:
$$\frac{2e^x}{e^{2x}+1}=\sum_{n=0}^\infty(-1)^ne^{(2n+1)x}$$
and we know:
$$e^x=\sum_{m=0}^\infty\frac{x^m}{m!}$$
now we can say:
$$\frac{2e^x}{e^{2x}+1}=\sum_{n=0}^\infty\sum_{m=0}^\infty\frac{(-1)^n\left[(2n+1)x\right]^m}{m!}$$
