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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?

user3141592
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1 Answers1

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

G Cab
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  • Sorry, but, what do $k$ and $t$ mean in these last expressions? I think that it should be $1 \le n$ in the summation operator, like the $n$ I used in my question, isn't it? And what does $t$ refer to? Is $t$ an arbitrary value or it depends on $k$? – user3141592 Jul 04 '17 at 18:55
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    @user3141592: added some clarification on the symbols adopted. $k$ is the summation index, going from $1$ on (then till to $n$, because for $n<k$ the binomial is null), thus exactly as the the sum you posted. $t$ is a free variable (as the usual $x$) and $\left[ {t^n } \right]$ is as in the added explanation. Conclusion: if you expand the indicated $f(t)$ in Taylor series, the the coefficient of $t^n$ is equal to the sum you posted. – G Cab Jul 04 '17 at 21:13
  • On the other hand, if I wanted to know the exact value of the sum, I would need to compute the Taylor series expansion of your expression, and it is more or less the same work. Wouldn't there be any closed form or any other way to simplify the sum? Someone deleted a comment where he said that we cloud express it as $f^{(n)} (0)$. Maybe that could help. By the way, thank you for your great answer. – user3141592 Jul 05 '17 at 21:41
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    @user3141592: a) agree that computationally is not a progress; b)algebraically, it is a step to understand its properties c) the hint $f^{(n)} (0)$ was most probably in the direction of the $\left[ {t^n } \right]$ above. d) I (and hopefully somebody else) will try and find a better "closed" formula, but it's not an easy task – G Cab Jul 05 '17 at 23:40
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    @user3141592: I have tried my best but could not reach to a "compact" expression for this sum. At the best, I could reach to other expressions always centered on a sum, not easier than the starting one. – G Cab Jul 11 '17 at 23:36
  • @G Thank you very much for your interest, it is highly appreciated. I will continue trying different things, and post a comment if I find something interesting – user3141592 Jul 12 '17 at 23:37
  • @user3141592: glad of your appreciation. It is an interesting question, would be nice to find a "compact" formula for that. But do you have any hint that leads you to think that there could be one? – G Cab Jul 13 '17 at 13:38
  • @G No. In fact, it is not "too" likely for it to have a closed form. By the way, I thought about Summation by Parts. Do we know anything about $\sum_{k=1}^n {n \choose k} B_{k+1}$? – user3141592 Jul 15 '17 at 16:54
  • @user3141592: summation by parts .. definitely worthy to explore. I added something about the sum, and will keep adding other features that might be of interest., when I find. Good work. – G Cab Jul 15 '17 at 21:29
  • On the other hand, partial summation may be useless, since sooner or later we might have to face a finite sum of the type $\sum \frac{B_{n+a}}{(n+b)!}$ for some integers $a, b$, and only the value of the infinite series is known. Maybe another possibility would be playing around with the binomial transform, but that also seems useless. – user3141592 Jul 15 '17 at 22:27
  • My new approach: since $$\sum^n_{k=1} {{n}\choose{k}} \frac {B_{k+1}} {(k+1)!} = \sum^\infty_{k=1} {{n}\choose{k}} \frac {B_{k+1}} {(k+1)!}$$ and $${n \choose k} =\int_{-\pi}^{\pi} e^{-ikt} (1+e^{it})^n dt$$ Then $$\sum^n_{k=1} {{n}\choose{k}} \frac {B_{k+1}} {(k+1)!} = \sum^\infty_{k=1} \int_{-\pi}^{\pi} e^{-ikt} (1+e^{it})^n dt \frac {B_{k+1}} {(k+1)!} = \int_{-\pi}^{\pi} (1+e^{it})^n \sum_{k=1}^\infty e^{-ikt} \frac {B_{k+1}} {(k+1)!} dt$$ Again $$\sum_{k=1}^\infty e^{-ikt} \frac {B_{k+1}} {(k+1)!} = e^{-it} (\frac {e^{-it}}{1-e^{-e^{-it}}} - 1 - \frac{e^{-it}}{2})$$ Am I wrong? – user3141592 Aug 25 '17 at 07:40
  • Woops, I forgot the $\frac{1}{2\pi}$ factor – user3141592 Aug 25 '17 at 08:23
  • The integral formulation (apart the missing factor) is very interesting, of course depending on the use you aim to do of it. – G Cab Aug 25 '17 at 20:57