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?
 A: 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.
