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

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?

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  • $\begingroup$ 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. $\endgroup$ – Ian Apr 15 at 18:57
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    $\begingroup$ You could try acceleration methods designed for alternating series, for instance Van Wijngaarden transformation. See also this article. $\endgroup$ – Jean-Claude Arbaut Apr 15 at 18:58
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    $\begingroup$ $\sum _{n=2}^{\infty } \frac{(-1)^n}{\log (n)}=\int_0^{\infty } \left(1+\left(-1+2^{1-x}\right) \zeta (x)\right) \, dx$ $\endgroup$ – Mariusz Iwaniuk Apr 15 at 20:17
  • $\begingroup$ 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 $\Delta(\frac{1}{\log n}) \sim \frac{-1}{n (\log n)^2}$. $\endgroup$ – Daniel Schepler Apr 15 at 20:36
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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.

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