Sum of Euler Numbers by Binomial Coefficients Vanishes

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

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

Corollary
Let $n \in \Z_{>0}$ be a (strictly) positive integer.

Then:

where $E_n$ denotes the $n$th Euler number.

If:


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

Then:

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:

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


 * $\paren {\dfrac {\paren {2 n}! } {0! \paren {2 n}!} } E_0 + \paren {\dfrac {\paren {2 n}! } {2! \paren {2 n - 2 }!} } E_2 + \paren {\dfrac {\paren {2 n}! } {4! \paren {2 n - 4 }!} } E_4 + \cdots + \paren {\dfrac {\paren {2 n}! } {\paren {2 n}! 0!} } E_{2 n} = 0$

But those coefficients are the binomial coefficients:
 * $\dbinom {2 n} 0 E_0 + \dbinom {2 n} 2 E_2 + \dbinom {2 n} 4 E_4 + \dbinom {2 n} 6 E_6 + \cdots + \dbinom {2 n} {2 n} E_{2 n} = 0$

Hence the result.

Also see

 * Sum of Bernoulli Numbers by Binomial Coefficients Vanishes