Asymptotic expansion of the sum $ \sum\limits_{k=1}^{n} \frac{\binom{n+1}{k} B_k}{ 3^k-1 } $ 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_{k=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 
\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 :  $\displaystyle  \sum_{k=1}^{n} \frac{\binom{n+1}{k}B_k} {N^k} = \frac{n+1}{N^{n+1}} \sum_{k=1}^{N-1} k^n -1$ 
We replace $N$ by $3^{p+1}$
$\displaystyle \sum_{k=1}^{n} \frac{\binom{n+1}{k}B_k} {3^{(p+1)k}} = \frac{n+1}{3^{(p+1)(n+1)}} \sum_{k=1}^{3^{p+1}-1} k^n -1$
That is to say
$\displaystyle a_n = \sum_{p=0}^{+\infty} \left(\frac{n+1}{3^{(p+1)(n+1)}} \sum_{k=1}^{3^{p+1}-1} k^n -1\right)$
Or
$\displaystyle a_n = \sum_{p=0}^{+\infty} \left(  \frac{n+1}{3^{p+1}} \sum_{k=1}^{3^{p+1}-1} \left( \frac{k}{3^{p+1}} \right)^n -1\right)$
If I come by your help, to answer this question, I will publish a new formula of Riemann zeta function that I find elegant.
Thank you in advance for your help.
 A: If we expand $\frac{1}{3^k-1}$ as a geometric series the problem boils down to estimating
$$ \sum_{k=1}^{n}\binom{n+1}{k}\frac{B_k}{3^{\eta k}} \tag{1}$$
and by Faulhaber's formula
$$ S_n(m)=\sum_{k=1}^{m}k^n = \frac{1}{n+1}\sum_{k=0}^n\binom{n+1}{k}B_k^+ m^{n+1-k} \tag{2} $$
so by choosing $m=3^{\eta}$ we get:
$$ (n+1) S_n(3^\eta) = (n+1)\left[1+2^n+\ldots+3^{\eta n}\right] = \sum_{k=0}^{n}\binom{n+1}{k}\frac{3^{\eta(n+1)}}{3^{\eta k}}B_k^+\tag{3} $$
and:
$$ \frac{n+1}{3^{\eta(n+1)}}\left[1+2^n+\ldots+3^{\eta n}\right] = 1+\frac{n+1}{3^\eta}+\sum_{k=1}^{n}\binom{n+1}{k}\frac{B_k}{3^{\eta k}} \tag{4}$$
from which:
$$ (1) = \sum_{k=1}^{n}\binom{n+1}{k}\frac{B_k}{3^{\eta k}}=-1+\frac{n+1}{3^{\eta(n+1)}}\left[1+2^n+3^n+\ldots+(3^\eta-1)^n\right]\tag{5} $$
and it is not difficult to finish from here. You may notice that
$$ \frac{1}{3^{p+1}}\sum_{k=1}^{3^{p+1}-1}\left(\frac{k}{3^{p+1}}\right)^n -\frac{1}{n+1}$$
is the error term of the rectangle method applied to the function $f(x)=x^n$ over the interval $(0,1)$, partitioned into $3^{p+1}$ equal intervals.
