Sum of Bernoulli Numbers by Power of Two and Binomial Coefficient

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Theorem

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

Then:

\(\ds \sum_{k \mathop = 1}^n \dbinom {2 n + 1} {2 k} 2^{2 k} B_{2 k}\) \(=\) \(\ds \binom {2 n + 1} 2 2^2 B_2 + \binom {2 n + 1} 4 2^4 B_4 + \binom {2 n + 1} 6 2^6 B_6 + \cdots\)
\(\ds \) \(=\) \(\ds 2 n\)

where $B_n$ denotes the $n$th Bernoulli number.


Proof 1

Let $B_k$ denote the kth Bernoulli number

Let $B_k \left(x\right)$ denote the kth Bernoulli polynomial


By Value of Odd Bernoulli Polynomial at One Half:

\(\ds B_{2 n + 1} \left({\frac 1 2}\right)\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \sum_{k \mathop = 0}^{2 n + 1} \binom {2 n + 1} k B_k \left({\frac 1 2}\right)^{2 n + 1 - k}\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds 2^{2 n + 1} \sum_{k \mathop = 0}^{2 n + 1} \binom {2 n + 1} k B_k \left({\frac 1 2}\right)^{2 n + 1 - k}\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \sum_{k \mathop = 0}^{2 n + 1} \binom {2 n + 1} k B_k \left({\frac 1 2}\right)^{-k}\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \sum_{k \mathop = 0}^{2 n + 1} \binom {2 n + 1} k B_k 2^k\) \(=\) \(\ds 0\)


By Odd Bernoulli Numbers Vanish:

\(\ds \sum_{k \mathop = 0}^{2 n + 1} \binom {2 n + 1} k B_k 2^k\) \(=\) \(\ds \left({2 n + 1}\right) 2 B_1 + \sum_{k \mathop = 0}^n \binom {2 n + 1} {2 k} B_{2 k} 2^{2 k}\)
\(\ds \) \(=\) \(\ds -\left({2 n + 1}\right) + \sum_{k \mathop = 0}^n \binom {2 n + 1 } {2 k} B_{2 k} 2^{2 k}\) $B_1 = -\frac 1 2$
\(\ds \leadsto \ \ \) \(\ds \sum_{k \mathop = 0}^n \binom {2 n + 1} {2 k} B_{2 k} 2^{2 k}\) \(=\) \(\ds 2 n + 1\)
\(\ds \leadsto \ \ \) \(\ds \sum_{k \mathop = 1}^n \binom {2 n + 1} {2 k} B_{2 k} 2^{2 k}\) \(=\) \(\ds 2 n + 1 - B_0\)
\(\ds \) \(=\) \(\ds 2 n\) $B_0 = 1$

$\blacksquare$


Proof 2

The proof proceeds by induction.

For all $n \in \Z_{> 0}$, let $P \left({n}\right)$ be the proposition:

$\displaystyle \sum_{k \mathop = 1}^n \dbinom {2 n + 1} {2 k} 2^{2 k} B_{2 k} = 2 n$


Basis for the Induction

$P \left({1}\right)$ is the case:

\(\ds \binom {2 \times 1 + 1} 2 2^2 B_2\) \(=\) \(\ds \frac {2 \times 3} 2 \times 2^2 \times \frac 1 6\) Binomial Coefficient with Two, Definition of Bernoulli Numbers
\(\ds \) \(=\) \(\ds 2\) after simplification
\(\ds \) \(=\) \(\ds 2 \times 1\)


Thus $P \left({1}\right)$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that, if $P \left({r}\right)$ is true, where $r \ge 2$, then it logically follows that $P \left({r + 1}\right)$ is true.


So this is the induction hypothesis:

$\displaystyle \sum_{k \mathop = 1}^r \dbinom {2 r + 1} {2 k} 2^{2 k} B_{2 k} = 2 r$


from which it is to be shown that:

$\displaystyle \sum_{k \mathop = 1}^{r + 1} \dbinom {2 \left({r + 1}\right) + 1} {2 k} 2^{2 k} B_{2 k} = 2 \left({r + 1}\right)$


Induction Step

This is the induction step:


\(\ds \sum_{k \mathop = 1}^{r + 1} \dbinom {2 \left({r + 1}\right) + 1} {2 k} 2^{2 k} B_{2 k}\) \(=\) \(\ds \sum_{k \mathop = 1}^r \dbinom {2 \left({r + 1}\right) + 1} {2 k} 2^{2 k} B_{2 k} + \dbinom {2 \left({r + 1}\right) + 1} {2 \left({r + 1}\right)} 2^{2 \left({r + 1}\right)} B_{2 \left({r + 1}\right)}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^r \dbinom {2 r + 3} {2 k} 2^{2 k} B_{2 k} + \left({2 r + 3}\right) 2^{2 r + 2} B_{2 r + 2}\) Binomial Coefficient with Self minus One and simplification



So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall n \in \Z_{>0}: \displaystyle \sum_{k \mathop = 1}^n \dbinom {2 n + 1} {2 k} 2^{2 k} B_{2 k} = 2 n$


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