How to prove that:
$$\sum_{k=0}^n \binom n k 2^k B_k = (2-2^n)B_n$$
In this sum, $B_n$ is the Bernoulli number with $B_1 = -\frac 1 2$. Thanks for your attention!
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John McClane
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Katy
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What have you tried? Please include your attempts and thoughts on the problem in the post. – Sil May 26 '19 at 19:06
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The identity that you want to verify can be rewritten as $$\sum^n_{k=0}\frac{2^k B_k}{k!(n-k)!} = 2\frac{B_n}{n!} - \frac{2^nB_n}{n!}$$ From the (standard) definition of Bernoulli numbers $B_n$, $g(z):=\frac{z}{e^z-1}=\sum^\infty_{n=0}\frac{B_n}{n!}z^n$. Hence, the right hand side of the first equation above corresponds to $n$-th coefficient of the power series \begin{align} f(z)&=2g(z) - g(2z)\\ &=2\frac{z}{e^z-1}-\frac{2z}{e^{2z}-1}\\ &=\frac{2z}{e^{2z} -1}e^z = g(2z)e^z\\ &=\Big(\sum^\infty_{n=0}\frac{B_n}{n!}2^nz^n\Big)\Big(\sum^\infty_{n=0}\frac{1}{n!}z^n\Big)\\ &=\sum^\infty_{n=0}c_nz^n \end{align} where $c_n=\sum^n_{k=0}\frac{2^kB_k}{k!}\frac{1}{(n-k)!}$ which is the left hands side of the identity we are trying to verify.
Mittens
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Could you please explain what I need to do next? I can't present this sum 2z*e^z/(e^(2z)-1) in series from n=0 to infinity. – Katy May 27 '19 at 17:17
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I know these definitions. However, I can't find Cn/n! from sum 2z*e^z/(e^(2z)-1) in series, because I can't present this sum in series from n=0 to infinity. – Katy May 27 '19 at 18:08
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There are no questions to this. How to find Cn? I want to find the coefficient, because it is the amount that I am trying to find. Therefore, I need to represent function 2z*e^z/(e^(2z)-1) as an infinite sum. – Katy May 27 '19 at 18:19
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$\frac{2ze^z}{e^{2z}-1}=\frac{2z}{e^{2z}-1}e^z=g(2z)e^z$ where $g(z)=\frac{z}{e^z-1}=\sum^\infty_{n=0}\frac{B_n}{n!}z^n$ – Mittens May 27 '19 at 18:21
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One possible way to demonstrate the thesis is through the B. Polynomials $$ \eqalign{ & S(n) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( \matrix{ n \cr k \cr} \right)2^{\,k} B_{\,k} } = 2^{\,n} \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( \matrix{ n \cr n - k \cr} \right)\left( {{1 \over 2}} \right)^{\,n - k} B_{\,k} } = \cr & = 2^{\,n} \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( \matrix{ n \cr k \cr} \right)\left( {{1 \over 2}} \right)^{\,k} B_{\,n - k} } = 2^{\,n} B_{\,n} (1/2) \cr} $$
It is known that the values of Bernoulli Polynomials at $x=1/2 are given by $$ B_{\,n} (1/2) = \left( {{1 \over {2^{\,n - 1} }} - 1} \right)B_{\,n} $$ and therefrom the demonstration that $$ S(n) = \left( {2 - 2^{\,n} } \right)B_{\,n} $$
The demonstration of the identity for $B_{\,n} (1/2)$ descends from the multiplicative identity $$ \eqalign{ & B_{\,n} (mx) = m^{\,n - 1} \sum\limits_{k = 0}^{m - 1} {B_{\,n} (x + k/m)} \quad \Rightarrow \cr & \Rightarrow \quad B_{\,n} \left( {2 \cdot {1 \over 2}} \right) = B_{\,n} \left( 1 \right) = 2^{\,n - 1} \left( {B_{\,n} \left( {{1 \over 2}} \right) + B_{\,n} \left( {{1 \over 2} + {1 \over 2}} \right)} \right)\quad \Rightarrow \cr & \Rightarrow \quad B_{\,n} \left( {{1 \over 2}} \right) = \left( {{1 \over {2^{\,n - 1} }} - 1} \right)B_{\,n} \left( 1 \right) = \left( {{1 \over {2^{\,n - 1} }} - 1} \right)B_{\,n} \left( 0 \right) = \cr & = \left( {{1 \over {2^{\,n - 1} }} - 1} \right)B_{\,n} \quad \left| {\;0 \le n} \right. \cr} $$
and you can find a proof of the multiplicative identity in this related post, which is not .. "very complicated".
G Cab
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1Ok, how i can prove that (i don't understand how to write formulas here, but i will try): $$B_n(1/2)=(1/(2^(n-1))-1)B_n&&? – Katy May 27 '19 at 14:52
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That requires to know things about Bernoulli polynomials...I still think that what you need to prove the identity requires only basic knowledge of product of analytic functions and the use of the definition of Bernoulli numbers $B_n$. In standard complex analysis textbooks, they are defined from the derivative at $0$ of the function by $f(z)=\frac{z}{e^z-1}$, that is $f(z)=\sum^\infty_{n=0}\frac{B_n}{n!}z^n$ – Mittens May 27 '19 at 17:55
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@Katy: added for proving the identity for $B_n(1/2)$, I wish that meets your expectations, and in any case it is good to brush it up for me and other readers here. – G Cab May 27 '19 at 20:54
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@OliverDiaz: you can, more politely, sustain the advantage of your answer inside its context. Yet criticism is always welcome. – G Cab May 27 '19 at 20:59
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Stop, @G Cab, why $B_n(0)=B_n(1)$? I know that this possible only when n not 1. – Katy May 27 '19 at 22:32
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@Katy: right, but for $n=1$ the factor in front $2^{(-n+1)}-1 $ is null, so the identity is true for all $n$ – G Cab May 28 '19 at 00:04
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