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 $\map P n$ be the proposition:

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


Basis for the Induction

$\map P 1$ 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 $\map P 1$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

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


So this is the induction hypothesis:

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


from which it is to be shown that:

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


Induction Step

This is the induction step:


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


This needs considerable tedious hard slog to complete it.
In particular: This approach may not be workable. May be easier / better to start with Definition:Bernoulli Numbers/Recurrence Relation and take it from there.
To discuss this page in more detail, feel free to use the talk page.
When this work has been completed, you may remove this instance of {{Finish}} from the code.
If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.


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


Therefore:

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


Also presented as

Sum of Bernoulli Numbers by Power of Two and Binomial Coefficient can also be presented using the archaic form of the Bernoulli numbers:

\(\ds \sum_{k \mathop = 1}^n \paren {-1}^{n - 1} \dbinom {2 n + 1} {2 k} 2^{2 k} {B_k}^*\) \(=\) \(\ds \binom {2 n + 1} 2 2^2 {B_1}^* - \binom {2 n + 1} 4 2^4 {B_2}^* + \binom {2 n + 1} 6 2^6 {B_3}^* - \cdots\)
\(\ds \) \(=\) \(\ds 2 n\)


Sources

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