Theorem: Sum of Euler Numbers by Binomial Coefficients Vanishes

$\forall n \in \Z_{>0}: \ds \sum_{k \mathop = 0}^n \binom {2 n} {2 k} E_{2 k} = 0$
where $E_k$ denotes the $k$th Euler number.

Proof:
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop = 0}^\infty \frac {E_n x^n} {n!}
      | r = \frac {2 e^x} {e^{2 x} + 1}
      | c = {{Defof|Euler Numbers}}
}}
{{eqn | r = \paren {\frac {2 e^x} {e^{2 x} + 1 } } \paren {\frac {e^{-x} } {e^{-x} } }
      | c = multiplying top and bottom by $e^{-x}$
}}
{{eqn | r = \paren {\frac 2 {e^x + e^{-x} } }
      | c = 
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = 1
      | r = \paren {\frac 2 {e^x + e^{-x} } } \paren {\frac {e^x + e^{-x} } 2}
      | c = 
}}
{{eqn | r = \paren {\sum_{n \mathop = 0}^\infty \frac {E_n x^n} {n!} } \paren {\sum_{n \mathop = 0}^\infty \frac {x^{2 n} } {\paren {2 n}!} }
      | c = {{Defof|Hyperbolic Cosine}} and Power Series Expansion for Hyperbolic Cosine Function
}}
{{end-eqn}}
Let:
{{begin-eqn}}
{{eqn | l = a_n
      | r = \frac {E_n x^n} {n!}
      | c = 
}}
{{eqn | l = b_n
      | r = \frac {x^{2 n} } {\paren {2 n}!} 
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop = 0}^\infty c_n
      | r = \paren {\sum_{n \mathop = 0}^\infty a_n} \paren {\sum_{n \mathop = 0}^\infty b_n}
      | rr= = 1
      | c = 
}}
{{eqn | l = c_n
      | r = \sum_{k \mathop = 0}^n a_k b_{n - k}
      | c = Product of Absolutely Convergent Series
}}
{{eqn | ll= \leadsto
      | l = c_0
      | r = \frac {E_0 x^0} {0!} \frac {x^0 } {0!} 
      | rr= = 1
      | c = as $c_0 = \paren {a_0} \paren {b_{0 - 0} } = \paren {a_0} \paren {b_0}$
}}
{{eqn | ll= \leadsto
      | l = \sum_{n \mathop = 1}^\infty c_n
      | r = \paren { \ds \sum_{n \mathop = 0}^\infty a_n } \paren {\ds \sum_{n \mathop = 0}^\infty b_n} - a_0 b_0
      | rr= = 0
      | c = subtracting $1$ from both sides
}}
{{end-eqn}}
We now have:
{{begin-eqn}}
{{eqn | l = c_1
      | r = \frac {E_0 x^0} {0!} \frac {x^2} {2!} + \frac {E_1 x^1} {1!} \frac {x^0} {0!}
      | c = $= a_0 b_1 + a_1 b_0$
}}
{{eqn | l = c_2
      | r = \frac {E_0 x^0} {0!} \frac {x^4} {4!} + \frac {E_1 x^1} {1!} \frac {x^2} {2!} + \frac {E_2 x^2} {2!} \frac {x^0} {0!}
      | c = $= a_0 b_2 + a_1 b_1 + a_2 b_0$
}}
{{eqn | l = c_3
      | r = \frac {E_0 x^0} {0!} \frac {x^6} {6!} + \frac {E_1 x^1} {1!} \frac {x^4} {4!} + \frac {E_2 x^2} {2!} \frac {x^2} {2!} + \frac {E_3 x^3} {3!} \frac {x^0 } {0!}
      | c = $= a_0 b_3 + a_1 b_2 + a_2 b_1 + a_3 b_0$
}}
{{eqn | l = c_4
      | r = \frac {E_0 x^0} {0!} \frac {x^8} {8!} + \frac {E_1 x^1} {1!} \frac {x^6} {6!} + \frac {E_2 x^2} {2!} \frac {x^4} {4!} + \frac {E_3 x^3} {3!} \frac {x^2 } {2!} + \frac {E_4 x^4} {4!} \frac {x^0} {0!}
      | c = $= a_0 b_4 + a_1 b_3 + a_2 b_2 + a_3 b_1 + a_4 b_0$
}}
{{eqn | o = \cdots
}}
{{eqn | l = c_n
      | r = \frac {E_0 x^0} {0!} \frac {x^{2 n} } {\paren {2 n}!} + \frac {E_1 x^1} {1!} \frac {x^{2 n - 2} } {\paren {2 n - 2}!} + \frac {E_2 x^2} {2!} \frac {x^{2 n - 4} } {\paren {2 n - 4 }!} + \cdots + \frac {E_n x^n} {n!} \frac {x^0} {0!}
      | c = 
}}
{{end-eqn}}
Grouping terms with even exponents produces:
{{begin-eqn}}
{{eqn | l = \paren {\frac 1 {0! 2!} } E_0 + \paren {\frac 1 {2! 0!} } E_2
      | r = 0
      | c = $x^2$ term from $c_1$ and $c_2$
}}
{{eqn | l = \paren {\frac 1 {0! 4!} } E_0 + \paren {\frac 1 {2! 2!} } E_2 + \paren {\frac 1 {4! 0!} } E_4
      | r = 0
      | c = $x^4$ term from $c_2$, $c_3$ and $c_4$
}}
{{eqn | o = \cdots
}}
{{eqn | l = \paren {\frac 1 {0! \paren {2 n}!} } E_0 + \paren {\frac 1 {2! \paren {2 n - 2 }!} } E_2 + \paren {\frac 1 {4! \paren {2 n - 4 }!} } E_4 + \cdots + \paren {\frac 1 {\paren {2 n}! 0!} } E_{2 n}
      | r = 0
      | c = $x^{2 n}$ term from $c_n$, $c_{n + 1} \cdots c_{2 n}$
}}
{{end-eqn}}
$\forall n \in \Z_{>0}$, 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.
{{qed}}