Let $B_k$ be the Bernoulli numbers. I'm looking for a proof for the following $$\sum_{k=0}^n\binom nk\frac{B_k}{n+2-k}=\frac{B_{n+1}}{n+1}$$ Could anyone help me, please.
proving that $\sum_{k=0}^n\binom nk\frac{B_k}{n+2-k}=\frac{B_{n+1}}{n+1}$
2
$\begingroup$
real-analysis
sequences-and-series
summation
closed-form
bernoulli-numbers
3 Answers
0
Note that your relation is only true for $n\ge1$. Since,
$$\binom nk\frac1{n+2-k}=\frac1{n+1}\binom{n+1}k-\frac1{(n+1)(n+2)}\binom{n+2}k$$
we have
$$\begin{align} \sum_{k=0}^n\binom nk\frac{B_k}{n+2-k}&=\frac1{n+1}\sum_{k=0}^n\binom{n+1}kB_k-\frac1{(n+1)(n+2)}\sum_{k=0}^n\binom{n+2}kB_k\\ &=\frac1{n+1}\sum_{k=0}^n\binom{n+1}kB_k-\frac1{(n+1)(n+2)}\left(\sum_{k=0}^{n+1}\binom{n+2}kB_k-\binom{n+2}{n+1}B_{n+1}\right)\\ &=\frac{B_{n+1}}{n+1} \end{align}$$
because $\sum_{k=0}^n\binom{n+1}kB_k$ and $\sum_{k=0}^{n+1}\binom{n+2}kB_k$ are zero for $n\ge1$.
2
We would like to prove $$ \sum_{k=0}^n\binom nk\frac{B_k}{n+2-k}=\frac{B_{n+1}}{n+1} \tag1 $$ which rewrites $$ \sum_{k=0}^n\frac{1}{(n-k)!(2+n-k)}\cdot \frac{B_k}{k!}=\frac{B_{n+1}}{(n+1)!} \tag2 $$ on the left hand side we recognize the coefficient of a general term of a Cauchy product of two power series, $$ f(x):=\sum_{k=0}^\infty\frac{x^n}{n!(2+n)}= \qquad g(x):=\sum_{k=0}^\infty\frac{B_n}{n!}x^n=\frac{x}{e^x-1} \tag3 $$ by integrating termwise $\displaystyle te^t=\sum_{k=0}^\infty\frac{t^{n+1}}{n!}$ between $0$ and $x$, one finds that $$ f(x)=\frac{1+e^x(-1+x)}{x^2} \tag4 $$ then one may conclude by observing that $$ f(x)g(x)=\frac1x\left(\frac{x}{e^x-1}-1+x\right).\tag5 $$
2
Using the explicit formula for the Bernoulli polynomials $$B_{n}\left(x\right)=\sum_{k=0}^{n}\dbinom{n}{k}B_{k}x^{n-k} $$ we can see that $$xB_{n}\left(x\right)=\sum_{k=0}^{n}\dbinom{n}{k}B_{k}x^{n-k+1} $$ so integrating both side we get $$\sum_{k=0}^{n}\dbinom{n}{k}\frac{B_{k}}{n-k+2}=\int_{0}^{1}xB_{n}\left(x\right)dx $$ and now using the well-known identity $$\int B_{n}\left(x\right)dx=\frac{B_{n+1}\left(x\right)}{n+1} $$ we get $$\sum_{k=0}^{n}\dbinom{n}{k}\frac{B_{k}}{n-k+2}=\frac{B_{n+1}\left(1\right)}{n+1}-\frac{B_{n+2}\left(1\right)-B_{n+2}\left(0\right)}{\left(n+1\right)\left(n+2\right)} $$ and since for $n>1 $ we have $B_{n}=B_{n}\left(1\right)=B_{n}\left(0\right),$ we finally get $$\sum_{k=0}^{n}\dbinom{n}{k}\frac{B_{k}}{n-k+2}=\color{red}{\frac{B_{n+1}}{n+1}}.$$