Very slow convergence of a particular series? I've read that
$$
\sum_{k=2}^{\infty} \frac{1}{k (\log k)^2} = 2.1097\ldots
$$
However when I compute the partial sums it looks like a lot of terms are needed to even get the first decimals right.
My question is : how many terms do we need to sum in order to get, say, the first $5$ digits right ?
 A: If you are satisfied with an approximate answer:
The function $t \mapsto \frac{1}{t(\log t)^2}$ is strictly decreasing on $(1,\infty)$, so for $n \geqslant 2$ we have
$$\int_{n+1}^\infty \frac{dt}{t(\log t)^2} < \sum_{k=n+1}^\infty \frac{1}{k(\log k)^2} < \int_n^\infty \frac{dt}{t(\log t)^2}.$$
Substituting $u = \log t$, we obtain
$$\int_x^\infty \frac{dt}{t(\log t)^2} = \int_{\log x}^\infty \frac{du}{u^2} = \frac{1}{\log x}.$$
To get within $\varepsilon$ of the sum of the series, one needs $\frac{1}{\log (n+1)} < \varepsilon$, or equivalently $n > e^{1/\varepsilon}-1$ at least, and $n > e^{1/\varepsilon}$ is certainly sufficient.
It is a little unclear what error we can accept to get the first five (significant) digits right, but the error must be $< 10^{-4}$, and an error of $5\cdot 10^{-5}$ will probably produce the first five correct digits. That means we need at least the terms to
$$k = \left\lfloor e^{10^4}\right\rfloor \approx 8.8\cdot 10^{4342},$$
and for the smaller error of $5\cdot 10^5$, we need the terms to
$$k = \left\lfloor e^{2\cdot 10^4}\right\rfloor \approx 7.756\cdot 10^{8685}.$$
There is no way to get the first five digits correct by summing the series naively within the lifetime of the universe using computers resembling the currently available ones.
To find the sum, one can approximate the series (after a reasonable number of initial terms) by the integral and estimate the error made by the approximation.
A: Following Daniel Fischer's suggestion, we can compare the sum to its integral using the Euler-Maclaurin sum formula.
Specifically, write
$$
\sum_{k=2}^{\infty} \frac{1}{k (\log k)^2}  = \sum_{k=2}^{m-1} \frac{1}{k (\log k)^2} + \sum_{k=m}^{\infty} \frac{1}{k (\log k)^2}
$$
and apply the Euler-Maclaurin formula to the tail sum $\sum_{k=m}^{\infty} \frac{1}{k (\log k)^2}$.
If we set $p = 4$ in the formula on the linked page and use the given bound on the error term $R$ we find that we only need to take $m = 135$ to get an answer correct to twenty digits after the decimal point.  The formula thus yields
$$
\sum_{k=2}^{\infty} \frac{1}{k (\log k)^2} \doteqdot 2.10974\ 28012\ 36891\ 97447.
$$
