The situation : I am looking for an asymptotic expansion of the sum $\displaystyle a_n=\sum_{k=1}^{n} \frac{\binom{n+1}{k} B_k}{ 3^k-1 } $ when $n \to \infty$. (The $ B_k $ are the Bernoulli numbers defined by $ \displaystyle \frac{z}{e^{z}-1}=\underset{n=0}{\overset{+\infty }{\sum }}\frac{B_{n}}{n!}z^{n}$).

Context : The initial problem was that I need to calculate a radius of convergence of a power series $\displaystyle \sum_{n=1}^{} a_n z^n $. I have almost tried everything to calculate this asymptotic expansion of the $a_n$, but to no avail.

The numerical test (computing) shows that $\displaystyle \lim_{n\to +\infty} \frac{a_{n+1}}{a_n} = 1$, that is, the convergence radius of the series is equal to $1$. But I can not analytically prove it.

My attempts to solve it :

$\displaystyle \large \begin{align*} a_n=\sum_{k=1}^{n} \frac{\binom{n+1}{k} B_k}{3^k-1 } &= \sum_{k=1}^{n} \frac{\binom{n+1}{k}B_k3^{-k}}{ 1- 3^{-k} } \\ &= \sum_{k=1}^{n} \binom{n+1}{k}B_k3^{-k} \sum_{p=0}^{+\infty}3^{-pk} \\ &= \sum_{p=0}^{+\infty} \sum_{k=1}^{n} \frac{\binom{n+1}{k}B_k} {3^{(p+1)k}} \\ \end{align*} $

Using the Faulhaber's formula : $\displaystyle \large \sum_{k=1}^{n} \frac{\binom{n+1}{k}B_k} {N^k} = \frac{n+1}{N^{n+1}} \sum_{k=1}^{N-1} k^n -1$

We replace $N$ by $3^{p+1}$

$\displaystyle \large \sum_{k=1}^{n} \frac{\binom{n+1}{k}B_k} {3^{(p+1)k}} = \frac{n+1}{3^{(p+1)(n+1)}} \sum_{k=1}^{3^{p+1}-1} k^n -1$

That is to say

$\displaystyle \large a_n = \sum_{p=0}^{+\infty} \left(\frac{n+1}{3^{(p+1)(n+1)}} \sum_{k=1}^{3^{p+1}-1} k^n -1\right)$


$\displaystyle \large a_n = \sum_{p=0}^{+\infty} \left( \frac{n+1}{3^{p+1}} \sum_{k=1}^{3^{p+1}-1} \left( \frac{k}{3^{p+1}} \right)^n -1\right)$

If I come by your help, to answer this question, I will publish a new formula of Riemann zeta function that I find elegant.

  • $\begingroup$ Just an idea, that perhaps you already tried. Calling $b_k=B_k/(3^k-1)$, then $a_{n+1}-a_n=b_{n+1}+\sum_{k=0}^{n}\binom{n+1}{k}b_{k+1}$. Perhaps it is easier to prove that the RHS divided by $a_n$ tends to zero. $\endgroup$
    – efs
    Commented Jun 27, 2017 at 0:23
  • $\begingroup$ Experimentally $a_n$ seems to be asymptotic to $-C\log(n)$ for some constant $C$ close to (equal to ?) $1$ $\endgroup$ Commented Nov 5, 2017 at 17:26

1 Answer 1


Let $n\in\mathbb{N}_{\ge 1}$, from the identity \begin{align} (k+1)^{n+1}-k^{n+1}=(n+1)k^n+\sum_{\ell=0}^{n-1}\binom{n+1}{\ell}k^{\ell} \end{align} we have \begin{align} a_n&=\sum_{p\ge 1}\left(\frac{1}{3^{p(n+1)}}\sum_{k=0}^{3^{p}-1}\left((k+1)^{n+1}-k^{n+1}-\sum_{\ell=0}^{n-1}\binom{n+1}{\ell}k^{\ell}\right)-1\right)\\ &=-\sum_{p\ge 1}\frac{1}{3^{p(n+1)}}\sum_{k=0}^{3^{p}-1}\sum_{\ell=0}^{n-1}\binom{n+1}{\ell}k^{\ell}. \end{align} Hence, \begin{align} -a_n&=\sum_{p\ge 1}3^{-pn}+\sum_{\ell=1}^{n-1}\binom{n+1}{\ell}\sum_{p\ge 1}\frac{1}{3^{p(n+1)}}\sum_{k=1}^{3^{p}-1}k^{\ell}\\ &\ll 1+\sum_{\ell=1}^{n-1}\binom{n+1}{\ell}\sum_{k\ge 1}k^{\ell}\sum_{p\ge \log(k+1)/\log 3}\frac{1}{3^{p(n+1)}}\\ &\ll 1+\sum_{\ell=1}^{n-1}\binom{n+1}{\ell}\sum_{k\ge 1}k^{\ell}3^{-(\lfloor\log(k+1)/\log 3\rfloor+1)(n+1)}\\ &\ll 1+\sum_{\ell=1}^{n-1}\binom{n+1}{\ell}\sum_{k\ge 1}\frac{k^{\ell}}{(k+1)^{n+1}}\\ &\ll 1+\sum_{k\ge 1}\left(\sum_{\ell=0}^{n+1}\binom{n+1}{\ell}\frac{k^{\ell}}{(k+1)^{n+1}}-\frac{(n+1)k^{n}}{(k+1)^{n+1}}-\frac{k^{n+1}}{(k+1)^{n+1}}\right)\\ &=1+\sum_{k\ge 1}\left(1-\frac{(n+1)k^{n}}{(k+1)^{n+1}}-\frac{k^{n+1}}{(k+1)^{n+1}}\right). \end{align} Namely, \begin{align} a_{n}&\ll 1+\sum_{k\ge 1}\left(1-\frac{k^{n}}{(k+1)^{n}}\frac{n+1+k}{k+1}\right)\\ &\ll n^2+\sum_{k\ge n^3}\left(1-\frac{k^{n}}{(k+1)^{n}}\left(1+\frac{n}{k}\right)\right). \end{align} Note that $$O\left(\frac{n}{k^2}\right)+\frac{1}{k} =\frac{1}{n}\log\left(1+\frac{n}{k}\right)\le \frac{1}{k}$$ for $k\ge n^2$. Thus \begin{align} a_{n}&\ll n^2+n\sum_{k\ge n^3}\left(1-\frac{k}{k+1}\left(1+\frac{n}{k}\right)^{\frac{1}{n}}\right)\ll n^2. \end{align}


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