Consider the numbers $$a\_n=\frac{1}{n+1}\sum\_{k=0}^{n}\frac{2^{k-1}\binom{n+1}{k}B\_k}{2^{s+k-1}-1}, \ n\geq0,$$ where $s\neq1;0;-1;-2;-3;...$ is a fixed real number, and the $B\_k$ are the Bernoulli numbers ($B\_1=-1/2$).
I am trying to show that the power series $\sum\_{n=0}^{+\infty} a\_n x^n$ converges. Numerical calculations indicate that its radius of convergence is $1$, but I have been unable to prove it analytically.



Convergence of a power series
Let $s > 1$. Using Faulhaber's formula and some well known properties of the Bernoulli numbers, we have $$a\_n = \frac{1}{n+1} \sum\_{k = 0}^n 2^{-s} \left(\sum\_{j = 0}^{+\infty} 2^{-(s+k-1) j}\right) \binom{n+1}{k} B\_k = \frac{2^{-s}}{n+1} \sum\_{j = 0}^{+\infty} 2^{-(s-1) j} \sum\_{k = 0}^n (2^j)^{-k} \binom{n+1}{k} B\_k = \frac{2^{-s}}{n+1} \sum\_{j = 0}^{+\infty} 2^{-(s-1) j} 2^{-(n+1)j} (B\_{n+1}(2^j)-B\_{n+1}) = 2^{-s} \sum\_{j = 0}^{+\infty} 2^{-(n+s)j} \sum\_{k = 0}^{2^j - 1} k^n.$$ Since $2^{-(n+s)j} k^n$ is non-negative, Fubini's theorem leads to $$a\_n = 2^{-s} \sum\_{k = 0}^{+\infty} k^n \sum\_{j = \lceil \log\_2(k+1)\rceil}^{+\infty} 2^{-(n+s)j} = 2^{-s} \sum\_{k = 0}^{+\infty} \frac{k^n}{2^{(s+n) \lceil \log\_2(k+1)\rceil}} \frac{1}{1-2^{-(s+n)}} = 2^{-s} \sum\_{k = 0}^{+\infty} b\_{k,s}(n).$$ We have the obvious inequality (using $\lceil x \rceil \geq x$) $b\_{k,s}(n) \leq \frac{1}{(k+1)^s (1-2^{-s})} (\*)$. Now, define $i \in \mathbb{N}$ such that $2^i \leq k \leq 2^{i+1}-1$, one has $$b\_{k,s}(n) = \frac{1}{2^{s(i+1)}(1-2^{-(s+n)})} \left(\frac{k}{2^{i+1}}\right)^n \underset{n \to +\infty}{\to} 0.$$ Using $(\*)$, we can apply the DCT which gives $a\_n \underset{n \to +\infty}{\to} 0$. This shows that the radius of convergence of your series is greater than $1$. On the other hand, we have (since $b\_{k,s}(n) \geq 0$) $$\sum\_{k = 0}^{+\infty} b\_{k,s}(n) \geq \sum\_{k = 0}^{+\infty} b\_{2^k-1,s}(n) = \sum\_{k = 0}^{+\infty} \frac{(2^k-1)^n}{1-2^{-(s+n)}} 2^{-(s+n)k} \geq \sum\_{k = 0}^{+\infty} \left(1-2^{-k}\right)^n 2^{-\lceil s \rceil k} = B\_n(\lceil s \rceil)$$. This $B\_n(j)$ is well known as a supertelescoping series (it is analogous to the asymptotic probability of a tie for first place). Indeed, one has for $j \geq 2$ $$(n+1) B\_n(j) \geq \sum\_{k=0}^{+\infty} 2^{-(j-1)(k+1)} \int\_{1-2^{-k}}^{1-2^{-(k+1)}} (n+1)x^{n} dx = \sum\_{k=0}^{+\infty} 2^{-(j-1)(k+1)} ((1-2^{-(k+1)})^{n+1} - (1-2^{-k})^{n+1}) = \sum\_{k=0}^{+\infty} (1-2^{-(j-1)})2^{-(j-1) (k+1)}(1-2^{-(k+1)})^{n+1} = (1-2^{-(j-1)}) B\_{n+1}(j-1).$$ By induction, we get $$B\_n(j) \geq B\_{n+j-1}(1) \prod\_{i = 1}^{j-1}\frac{1-2^{-i}}{n+i}.$$ Since $B\_{n}(1) \geq \sum\_{k=0}^{+\infty} ((1-2^{-(k+1)})^{n+1} - (1-2^{-k})^{n+1}) = 1$, we obtain $a\_n \geq 2^{-s} \prod\_{i = 1}^{\lceil s \rceil -1}\frac{1-2^{-i}}{n+i} \geq \frac{C}{n^{\lceil s \rceil}}$ for some $C > 0$ and thus the radius of convergence of your series is exactly $1$.


