Find the infinite sum $\sum^\infty_{n=2}\frac{\left(-1\right)^{n}}{\log n}$, speed up its convergence

$$\displaystyle\sum_{n\in\mathbb{Z}_{\geqslant{2}}}\frac{\left(-1\right)^{n}}{\log n}$$

Since $$\log x$$ grows more slowly than any positive power, the sum above converges extremely slowly. Is there a series that converges faster and converges to the same value as this sum? Or an integral?

• You could use the Euler-Maclaurin formula with a suitable smooth, integrable interpolant. Of course that requires the value of the integral of the interpolant. – Ian Apr 15 at 18:57
• You could try acceleration methods designed for alternating series, for instance Van Wijngaarden transformation. See also this article. – Jean-Claude Arbaut Apr 15 at 18:58
• $\sum _{n=2}^{\infty } \frac{(-1)^n}{\log (n)}=\int_0^{\infty } \left(1+\left(-1+2^{1-x}\right) \zeta (x)\right) \, dx$ – Mariusz Iwaniuk Apr 15 at 20:17
• Summation by parts might work to accelerate convergence a bit - it would express the sum in terms of $\sum_{n=2}^\infty [\frac{1}{2} (-1)^n] \Delta(\frac{1}{\log n})$ where $\Delta(\frac{1}{\log n}) \sim \frac{-1}{n (\log n)^2}$. – Daniel Schepler Apr 15 at 20:36

1 Answer

A first transformation that gives an equivalent series which converges a it more rapidly is to break the series into even and odd terms, and let $$n = 2k$$ or $$2k+1$$. then you get $$\sum_{k=1}^\infty \left( \frac{1}{\log (2k)} - \frac{1}{\log (2k+1)} \right) = \sum_{k=1}^\infty \frac{\log\left( 1+\frac 1{2k}\right)}{\log(2k)\log(2k+1)}$$ And now you can work with that because $$\log\left( 1+\frac 1{2k}\right) = \sum_1^\infty \frac{(-1)^{m+1}}{(m+1)(2k)^m}$$ $$\sum_{k=1}^\infty \frac{\log\left( 1+\frac 1{2k}\right)}{\log(2k)\log(2k+1)} = \sum_{k=1}^\infty \frac{\sum_{m=1}^\infty \frac{(-1)^{m+1}}{(m+1)(2k)^m}}{\log(2k)\log(2k+1)} = \sum_{m=1}^\infty \frac{(-1)^{m+1}}{m+1}\sum_{k=1}^\infty \frac{ 1}{(2k)^m\log(2k)\log(2k+1)}$$ Now we see another alternating sum, in $$m$$, and we can apply the same sort of technique, but the algebra gets a bit messy if we do. However, when $$m > 1$$ the sum over $$k$$ converges reasonably rapidly due to the power of $$k^m$$ in the denominator. So we are left with just one headache, that is how to quickly approximate $$\sum_{k=1}^\infty \frac{ 1}{(2k)\log(2k)\log(2k+1)}$$ That is predictably slow to converge, since with9out the logs it would diverge and the logs grow slowly. However, it does converge much more rapidly than the original sum.