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.

  • 1
    $\begingroup$ You can use the same argument in this link mathoverflow.net/questions/273001/… to demonstrate that the radius of convergence of your series is greater than 1. You just need to lower bound $|a_n|$ by a sequence $b_n$ such that the radius of convergence of $\sum b_n x^n$ is equal to 1. $\endgroup$
    – Pascal
    Jul 14, 2023 at 23:27

1 Answer 1


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 $1$.


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