Evaluate $\lim\limits_{n\to\infty}(f(n+1)-f(n))$ where $f(n)=|B_{2n}|^{1/2n}$

Question

Let $f(n)=|B_{2n}|^{1/2n}$ where $B_{2n}$ is the $(2n)$-th Bernoulli's number. Evaluate the limit $$\lim\limits_{n\to\infty}(f(n+1)-f(n)).$$

How can I calculate this limit?

Answer 1

Since $B_{2n} =\dfrac{(-1)^{n+1}2(2n)!}{(2\pi)^{2n}}\zeta(2n) $ and $n! =\approx \sqrt{2\pi}\dfrac{n^{n+1/2}}{e^n}e^{1/(12n)+O(1/n^3)} $,

$\begin{array}\\ B_{2n} &=\dfrac{(-1)^{n+1}2(2n)!}{(2\pi)^{2n}}\zeta(2n)\\ &=\dfrac{(-1)^{n+1}2\sqrt{4\pi n}\dfrac{(2n)^{2n}}{e^{2n}}e^{1/(24n)+O(1/n^3)}}{(2\pi)^{2n}}(1+O(2^{-2n}))\\ \text{so}\\ f(n) &=\dfrac{(2\sqrt{4\pi n})^{1/(2n)}\dfrac{2n}{e}e^{1/(48n^2)+O(1/n^4)}}{2\pi}(1+O(2^{-2n}/n))\\ &=\dfrac{(2\sqrt{4\pi n})^{1/(2n)}ne^{1/(48n^2)+O(1/n^4)}}{\pi e}(1+O(2^{-2n}/n))\\ &=\dfrac{n}{\pi e}(2\sqrt{4\pi n})^{1/(2n)}e^{1/(48n^2)+O(1/n^4)}(1+O(2^{-2n}/n))\\ &=\dfrac{n}{\pi e}(1+\dfrac{\ln(16\pi n)}{4n}+O(\ln^2(n)/n^2))(1+1/(48n^2)+O(1/n^4))(1+O(2^{-2n}/n))\\ &=\dfrac{n}{\pi e}+\dfrac{\ln(16\pi n)}{4\pi e}+O(\ln^2(n)/n)\\ &=\dfrac{n}{\pi e}+\dfrac{\ln(n)}{4\pi e}+\dfrac{\ln(16\pi )}{4\pi e}+O(\ln^2(n)/n)\\ \text{so}\\ f(n)-f(n-1) &=\dfrac{1}{\pi e}+\dfrac{1}{4n\pi e}+o(1)\\ \end{array} $

Answer 2

Hint. An asymptotic approximation of the Bernoulli numbers can be found HERE (see formula (40)). It can be obtained from the Stirling approximation $n!\sim\sqrt{2\pi n}(n/e)^n$: $$|B_{2n}| =\frac{2(2n)!}{(2\pi)^{2n}}\zeta(2n)\sim 2\sqrt{2\pi(2n)}\,\left(\frac {2n}{2\pi\,e}\right)^{2n}\implies f(n)=|B_{2n}|^{1/(2n)}\sim \frac n{\pi\,e}.$$ By using a more precise variant of the above approximation for the factorial, you should be able to evaluate the limit of $f(n+1)-f(n)$.