# Computing a slowly-converging limit

Let $$f(x)=-\log\log x+\sum_{2\le n\le x}\frac{1}{n\log n}.$$

How can I efficiently compute $$f(\infty)=:\lim_{x\to\infty}f(x)?$$

Brute force suffices to find 0.7946786454... but I would like several hundred digits.

It seems that I should be able to use numerical integration, since $$\frac{1}{n\log n}-\log\log n+\log\log(n-1)$$ is smooth (and appears to be monotonic). (In fact, it even has a closed-form integral in li.) Alternately, various forms of series acceleration may apply.

(N.B. I have no real experience with numerical analysis outside an undergraduate class a few years back.)

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The short almost-answer: Look at the Euler–Maclaurin formula and apply it to $\int 1/(x\log x)\,dx=\log\log x$. Use it to get good approximations to tails of the series. – Harald Hanche-Olsen Nov 9 '12 at 17:08