How can we go about proving: As $n \to \infty$
$$\dfrac{4}{\pi}\sum_{k = 1}^{\infty}\dfrac{(-1)^{k - 1}k}{B_{2k}(2k - 1)}\left(\dfrac{\log n}{2\pi}\right)^{2k - 1} \approx \dfrac{n}{\text{ln} \text{ } n}\tag 1\label{1}$$
where $B_m$ denotes the $m$-th Bernoulli number for all $m \geqslant 2$. I have tried this problem for quite considerable amount of time. Indeed, I was able to simplify the LHS more using some clever substitutions and some analytic number theory.
I was able to prove that: As $n \to \infty$
$$\int_{0}^{\infty}\dfrac{(\log n)^\tau d\tau}{\Gamma(\tau + 1)\zeta(\tau + 1)\tau} \approx \dfrac{4}{\pi}\sum_{k = 1}^{\infty}\dfrac{(-1)^{k - 1}k}{B_{2k}(2k - 1)}\left(\dfrac{\log n}{2\pi}\right)^{2k - 1} \tag 2\label{2}$$
however I'm not quite satisfied with my proof for the same. I am wondering if there's a nice way to prove $\eqref{1}$ and $\eqref{2}$ that doesn't make use of complex analysis. Your help for proving $\eqref{2}$ would be highly appreciated. I'm looking for a detailed real analytic answer.
Your help would be highly appreciated. Thanks.
No idea of what you are trying to do.
The asymptotic of Bernouilli numbers can be obtained directly from that $\frac{x}{e^x-1}-\frac{2i\pi}{x-2i\pi}-\frac{-2i\pi}{x+2i\pi}$ is analytic on a larger disk $|x|<4\pi$, but for clarity it is convenient to know that
$$\zeta(2k) = (-1)^{k+1}\frac{(2\pi)^{2k}B_{2k}}{2(2k)!} = 1+O(2^{-2k}))$$
Which gives that
$$\sum_{k \ge 1} \frac{2(-1)^{k+1}}{(2\pi)^{2k}B_{2k}} x^{2k}\frac{k}{2k-1}=\sum_{k\ge 1} \frac{x^{2k}}{\zeta(2k) (2k)!}\frac{k}{2k-1}=\sum_{k\ge 1} \frac{x^{2k}}{ (2k)!}\frac{k}{2k-1} (1+O(2^{-2k})$$ as $x\to \infty$ it is $$=\frac14 e^x+O(e^x/x)+O(e^{x/2})$$