Series involving Bernoulli Numbers

Question

I would be interested in the following sum:

$$\sum^n_{k=1} {{n}\choose{k}} \frac {B_{k+1}} {(k+1)!}$$

It would be great to get a closed form for it. The problem is that I do not know what to do with those Bernoulli Numbers. I have tried to relate this series to others containing these numbers, such as the Taylor series of the arctangent function, sum of powers of consecutive integers or the famous $\frac{t}{e^t-1}$ series expansion, with no result.

Could anyone give me any hint about how to evaluate this sum? By the way, is there a general procedure for working with series involving Bernoulli Numbers?

Answer 1

Since, by definition of Bernoulli Numbers $$ {t \over {e^{\,t} - 1}} = \sum\limits_{0\, \le \,k} {{{B_{\,k} } \over {k!}}t^{\,k} } \quad \Rightarrow \quad {1 \over {e^{\,t} - 1}} - {1 \over t} = \sum\limits_{0\, \le \,k} {{{B_{\,k + 1} } \over {\left( {k + 1} \right)!}}t^{\,k} } = - {1 \over 2} + \sum\limits_{1\, \le \,k} {{{B_{\,k + 1} } \over {\left( {k + 1} \right)!}}t^{\,k} } $$ and since $$ \left( {1 + t} \right)^n = \sum\limits_{0\, \le \,j\,\left( { \le \,n} \right)} {\left( \matrix{ n \cr j \cr} \right)t^{\,j} } $$ then multiplying the two $$ \eqalign{ & \left( {1 + t} \right)^n \left( {{1 \over {e^{\,t} - 1}} - {1 \over t} + {1 \over 2}} \right) = \sum\limits_{0\, \le \,j\,\left( { \le \,n} \right)} {\left( \matrix{ n \cr j \cr} \right)t^{\,j} \sum\limits_{1\, \le \,k} {{{B_{\,k + 1} } \over {\left( {k + 1} \right)!}}t^{\,k} } } = \cr & = \sum\limits_{1\, \le \,s\,} {\left( {\sum\limits_{1\, \le \,k} {\left( \matrix{ n \cr s - k \cr} \right){{B_{\,k + 1} } \over {\left( {k + 1} \right)!}}} } \right)t^{\,s} } \cr} $$ which means that $$ \eqalign{ & \sum\limits_{1\, \le \,k} {\left( \matrix{ n \cr n - k \cr} \right){{B_{\,k + 1} } \over {\left( {k + 1} \right)!}}} = \sum\limits_{1\, \le \,k} {\left( \matrix{ n \cr k \cr} \right){{B_{\,k + 1} } \over {\left( {k + 1} \right)!}}} = \cr & = \left[ {t^n } \right]\left( {1 + t} \right)^n \left( {{1 \over {e^{\,t} - 1}} - {1 \over t} + {1 \over 2}} \right) = \cr & = \left[ {t^n } \right]\left( {1 + t} \right)^n \left( {{{2t + \left( {t - 2} \right)\left( {e^{\,t} - 1} \right)} \over {2t\left( {e^{\,t} - 1} \right)}}} \right) \cr} $$ as rightly indicated by Marko Riedel,
and where:

• $t$ is a variable
• $\left[ {t^n } \right]$ is a standard form to indicate the factor multiplying $t^n$ in the Taylor expansion of the following function, i.e. $1/n! f^{(n)}(0)$

Addendum

Concerning your comment requesting about $\sum\limits_{1\, \le \,k\, \le \,n} {\left( \matrix{ n \cr k \cr} \right)B_{\,k + 1} } $, starting from: $$ \eqalign{ & \left[ {1 = n} \right] = \sum\limits_{0\, \le \,k\, \le \,n - 1} {\left( \matrix{ n \cr k \cr} \right)B_{\,k} } = \sum\limits_{0\, \le \,k\, \le \,n - 1} {\left( \matrix{ n - 1 \cr k \cr} \right)B_{\,k} } + \sum\limits_{0\, \le \,k\, \le \,n - 1} {\left( \matrix{ n - 1 \cr k - 1 \cr} \right)B_{\,k} } = \cr & = \sum\limits_{0\, \le \,k\, \le \,n - 1} {\left( \matrix{ n - 1 \cr k \cr} \right)B_{\,k} } + \sum\limits_{0\, \le \,k\, \le \,n - 2} {\left( \matrix{ n - 1 \cr k \cr} \right)B_{\,k + 1} } = \cr & = \sum\limits_{0\, \le \,k\, \le \,n - 2} {\left( \matrix{ n - 1 \cr k \cr} \right)B_{\,k} } + B_{\,n - 1} + B_{\,1} + \sum\limits_{1\, \le \,k\, \le \,n - 1} {\left( \matrix{ n - 1 \cr k \cr} \right)B_{\,k + 1} } - B_{\,n} = \cr & = \left[ {n = 2} \right] + B_{\,n - 1} + B_{\,1} + \sum\limits_{1\, \le \,k\, \le \,n - 1} {\left( \matrix{ n - 1 \cr k \cr} \right)B_{\,k + 1} } - B_{\,n} \cr} $$ (where $[P]$ denotes the Iverson bracket)

then we have $$ \sum\limits_{1\, \le \,k\, \le \,n} {\left( \matrix{ n \cr k \cr} \right)B_{\,k + 1} } = \left[ {0 = n} \right] - \left[ {n = 1} \right] - B_{\,1} + B_{\,n + 1} - B_{\,n} $$

Also, indicating by $b_n(x)$ the bernoulli polynomials $$ b_{\,n} (x) = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,n} \right)} {\left( \matrix{ n \cr j \cr} \right)B_{\,n - j} \;x^{\,j} } = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,n} \right)} {\left( \matrix{ n \cr j \cr} \right)b_{\,n - j} (0)\;x^{\,j} } = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,n} \right)} {\left( \matrix{ n \cr j \cr} \right)B_{\,j} \;x^{\,n - j} } $$ and $$ {{b_{\,n} (x)} \over {x^{\,n} }} = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,n} \right)} {\left( \matrix{ n \cr j \cr} \right)B_{\,j} \;\left( {{1 \over x}} \right)^{\,j} } \quad \Rightarrow \quad x^{\,n} b_{\,n} (1/x) = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,n} \right)} {\left( \matrix{ n \cr j \cr} \right)B_{\,j} \;x^{\,j} } $$ proceeding to split the binomial $$ \eqalign{ & x^{\,n} b_{\,n} (1/x) = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,n} \right)} {\left( \matrix{ n \cr j \cr} \right)B_{\,j} \;x^{\,j} } = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,n} \right)} {\left( \matrix{ n - 1 \cr j \cr} \right)B_{\,j} \;x^{\,j} } + \sum\limits_{\left( {1\, \le } \right)\,j\,\left( { \le \,n} \right)} {\left( \matrix{ n - 1 \cr j - 1 \cr} \right)B_{\,j} \;x^{\,j} } = \cr & = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,n - 1} \right)} {\left( \matrix{ n - 1 \cr j \cr} \right)B_{\,j} \;x^{\,j} } + x\sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,n - 1} \right)} {\left( \matrix{ n - 1 \cr j \cr} \right)B_{\,j + 1} \;x^{\,j} } = \cr & = x^{\,n - 1} b_{\,n - 1} (1/x) + x\sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,n - 1} \right)} {\left( \matrix{ n - 1 \cr j \cr} \right)B_{\,j + 1} \;x^{\,j} } \cr} $$ so that $$ \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,n} \right)} {\left( \matrix{ n \cr j \cr} \right)B_{\,j + 1} \;x^{\,j} } = B_{\,1} \; + \sum\limits_{1\, \le \,j\,\,\left( { \le \,n} \right)} {\left( \matrix{ n \cr j \cr} \right)B_{\,j + 1} \;x^{\,j} } = {1 \over x}\left( {x^{\,n + 1} b_{\,n + 1} (1/x) - x^{\,n} b_{\,n} (1/x)} \right) $$