Sum of Euler Numbers by Binomial Coefficients Vanishes

Theorem
$\forall n \in \Z_{>0}: \displaystyle \sum_{k \mathop = 0}^{n} \binom {2n} {2k} E_{2k } = 0$

where $E_k$ denotes the $k$th Euler number.

Proof
Take the definition of Euler numbers:

From the definition of the exponential function:

Thus:

By Product of Absolutely Convergent Series, we will let:

Then:

We now have:

Grouping terms with even exponents produces:

$\forall n \in \Z_{\gt 0}$, multiplying the coefficients of $x^{2n}$ through by $\paren {2n }!$ gives:

But those coefficients are the binomial coefficients:

Hence the result.

Also see

 * Sum of Bernoulli Numbers by Binomial Coefficients Vanishes