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 \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
$\displaystyle \large a_n = \sum_{p=0}^{+\infty} \left(\frac{n+1}{3^{(p+1)(n+1)}} \sum_{k=1}^{3^{p+1}-1} k^n -1\right)$
Or
$\displaystyle \large a_n = \sum_{p=0}^{+\infty} \left(  \frac{n+1}{3^{p+1}} \sum_{k=1}^{3^{p+1}-1} \left( \frac{k}{3^{p+1}} \right)^n -1\right)$
If I come by your help, to answer this question, I will publish a new formula of Riemann zeta function that I find elegant.   
 A: 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}
