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:

\(\ds E_{2 n}\)

\(=\)

\(\ds -\sum_{k \mathop = 0}^{n - 1} \dbinom {2 n} {2 k} E_{2 k}\)

\(\ds \)

\(=\)

\(\ds -\paren {\binom {2 n} 0 E_0 + \binom {2 n} 2 E_2 + \binom {2 n} 4 E_4 + \cdots + \binom {2 n} {2 n - 2} E_{2 n - 2} }\)

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:

\(\ds \sum_{k \mathop = 0}^{n - 1} \dbinom {2 n} {2 k} E_{2 k} + E_{2 n}\)

\(=\)

\(\ds 0\)

\(\ds E_{2 n}\)

\(=\)

\(\ds - \sum_{k \mathop = 0}^{n - 1} \dbinom {2 n} {2 k} E_{2 k}\)

$\blacksquare$

Proof

Take the definition of Euler numbers:

\(\ds \sum_{n \mathop = 0}^\infty \frac {E_n x^n} {n!}\)

\(=\)

\(\ds \frac {2 e^x} {e^{2 x} + 1}\)

\(\ds \)

\(=\)

\(\ds \paren {\frac {2 e^x} {e^{2 x} + 1 } } \paren {\frac {e^{-x} } {e^{-x} } }\)

Multiply by $1$

\(\ds \)

\(=\)

\(\ds \paren {\frac 2 {e^x + e^{-x} } }\)

From the definition of the exponential function:

\(\ds e^x\)

\(=\)

\(\ds \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}\)

\(\ds \)

\(=\)

\(\ds 1 + x + \frac {x^2} {2!} + \frac {x^3} {3!} + \frac {x^4} {4!} + \cdots\)

\(\ds e^{-x}\)

\(=\)

\(\ds \sum_{n \mathop = 0}^\infty \frac {\paren {-x}^n} {n!}\)

\(\ds \)

\(=\)

\(\ds 1 - x + \frac {x^2} {2!} - \frac {x^3} {3!} + \frac {x^4} {4!} - \cdots\)

\(\ds \paren {\frac {e^x + e^{-x} } 2}\)

\(=\)

\(\ds \paren {\sum_{n \mathop = 0}^\infty \frac {x^{2 n} } {\paren {2 n}!} }\)

\(\ds \)

\(=\)

\(\ds 1 + \frac {x^2} {2!} + \frac {x^4} {4!} + \cdots\)

odd terms cancel in the sum.

Thus:

\(\ds 1\)

\(=\)

\(\ds \paren {\frac 2 {e^x + e^{-x} } } \paren {\frac {e^x + e^{-x} } 2}\)

\(\ds \)

\(=\)

\(\ds \paren {\sum_{n \mathop = 0}^\infty \frac {E_n x^n} {n!} } \paren {\sum_{n \mathop = 0}^\infty \frac {x^{2 n} } {\paren {2 n}!} }\)

By Product of Absolutely Convergent Series, we will let:

\(\ds a_n\)

\(=\)

\(\ds \frac {E_n x^n} {n!}\)

\(\ds b_n\)

\(=\)

\(\ds \frac {x^{2 n} } {\paren {2 n}!}\)

Then:

\(\ds \sum_{n \mathop = 0}^\infty c_n\)

\(=\)

\(\ds \paren {\sum_{n \mathop = 0}^\infty a_n} \paren {\sum_{n \mathop = 0}^\infty b_n}\)

\(\ds = 1\)

\(\ds c_n\)

\(=\)

\(\ds \sum_{k \mathop = 0}^n a_k b_{n - k}\)

\(\ds c_0\)

\(=\)

\(\ds \frac {E_0 x^0} {0!} \frac {x^0 } {0!}\)

\(\ds = 1\)

$c_0 = \paren {a_0} \paren {b_{0 - 0} } = \paren {a_0} \paren {b_0}$

\(\ds \leadsto \ \ \)

\(\ds \sum_{n \mathop = 1}^\infty c_n\)

\(=\)

\(\ds \paren { \displaystyle \sum_{n \mathop = 0}^\infty a_n } \paren {\displaystyle \sum_{n \mathop = 0}^\infty b_n} - a_0 b_0\)

\(\ds = 0\)

Subtract $1$ from both sides

We now have:

\(\ds c_1\)

\(=\)

\(\ds \frac {E_0 x^0} {0!} \frac {x^2} {2!} + \frac {E_1 x^1} {1!} \frac {x^0} {0!}\)

$= a_0 b_1 + a_1 b_0$

\(\ds c_2\)

\(=\)

\(\ds \frac {E_0 x^0} {0!} \frac {x^4} {4!} + \frac {E_1 x^1} {1!} \frac {x^2} {2!} + \frac {E_2 x^2} {2!} \frac {x^0} {0!}\)

$= a_0 b_2 + a_1 b_1 + a_2 b_0$

\(\ds c_3\)

\(=\)

\(\ds \frac {E_0 x^0} {0!} \frac {x^6} {6!} + \frac {E_1 x^1} {1!} \frac {x^4} {4!} + \frac {E_2 x^2} {2!} \frac {x^2} {2!} + \frac {E_3 x^3} {3!} \frac {x^0 } {0!}\)

$= a_0 b_3 + a_1 b_2 + a_2 b_1 + a_3 b_0$

\(\ds c_4\)

\(=\)

\(\ds \frac {E_0 x^0} {0!} \frac {x^8} {8!} + \frac {E_1 x^1} {1!} \frac {x^6} {6!} + \frac {E_2 x^2} {2!} \frac {x^4} {4!} + \frac {E_3 x^3} {3!} \frac {x^2 } {2!} + \frac {E_4 x^4} {4!} \frac {x^0} {0!}\)

$= a_0 b_4 + a_1 b_3 + a_2 b_2 + a_3 b_1 + a_4 b_0$

\(\ds \)

\(\cdots\)

\(\ds \)

\(\ds c_n\)

\(=\)

\(\ds \frac {E_0 x^0} {0!} \frac {x^{2 n} } {\paren {2 n}!} + \frac {E_1 x^1} {1!} \frac {x^{2 n - 2} } {\paren {2 n - 2}!} + \frac {E_2 x^2} {2!} \frac {x^{2 n - 4} } {\paren {2 n - 4 }!} + \cdots + \frac {E_n x^n} {n!} \frac {x^0} {0!}\)

Grouping terms with even exponents produces:

\(\ds \paren {\frac 1 {0! 2!} } E_0 + \paren {\frac 1 {2! 0!} } E_2\)

\(=\)

\(\ds 0\)

$x^2$ term from $c_1$ and $c_2$

\(\ds \paren {\frac 1 {0! 4!} } E_0 + \paren {\frac 1 {2! 2!} } E_2 + \paren {\frac 1 {4! 0!} } E_4\)

\(=\)

\(\ds 0\)

$x^4$ term from $c_2$, $c_3$ and $c_4$

\(\ds \)

\(\cdots\)

\(\ds \)

\(\ds \paren {\frac 1 {0! \paren {2 n}!} } E_0 + \paren {\frac 1 {2! \paren {2 n - 2 }!} } E_2 + \paren {\frac 1 {4! \paren {2 n - 4 }!} } E_4 + \cdots + \paren {\frac 1 {\paren {2 n}! 0!} } E_{2 n}\)

\(=\)

\(\ds 0\)

$x^{2 n}$ term from $c_n$, $c_{n + 1} \cdots c_{2 n}$

$\forall n \in \Z_{>0}$, 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.

$\blacksquare$

Examples

$\begin{array}{r|cccccccccc} E_k & \dbinom n 0 & & \dbinom n 2 & & \dbinom n 4 & & \dbinom n 6 & & \dbinom n 8 & & \dbinom n {10} \\ \hline E_0 = +1 & 1 E_0 & & & & & & & & & & & = 1 \\ E_2 = -1 & 1 E_0 & + & 1 E_2 & & & & & & & & & = 0 \\ E_4 = +5 & 1 E_0 & + & 6 E_2 & + & 1 E_4 & & & & & & & = 0 \\ E_6 = -61 & 1 E_0 & + & 15 E_2 & + & 15 E_4 & + & 1 E_6 & & & & & = 0 \\ E_8 = +1385 & 1 E_0 & + & 28 E_2 & + & 70 E_4 & + & 28 E_6 & + & 1 E_8 & & & = 0 \\ E_{10} = -50521 & 1 E_0 & + & 45 E_2 & + & 210 E_4 & + & 210 E_6 & + & 45 E_8 & + & 1 E_{10} & = 0 \\ \end{array}$

Also see

• Sum of Bernoulli Numbers by Binomial Coefficients Vanishes