Series involving Bernoulli Numbers 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?
 A: 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
$$
\sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,n} \right)} {\left( \matrix{
  n \cr 
  j \cr}  \right)B_{\,j + 1} \;x^{\,j} }  = B_{\,1} \; + \sum\limits_{1\, \le \,j\,\,\left( { \le \,n} \right)} {\left( \matrix{
  n \cr 
  j \cr}  \right)B_{\,j + 1} \;x^{\,j} }  = {1 \over x}\left( {x^{\,n + 1} b_{\,n + 1} (1/x) - x^{\,n} b_{\,n} (1/x)} \right)
$$
