While testing implementations of Wynn's $\epsilon$-algorithm and Levin's u-transformation I need the value of $$\sum_{n=0}^{\infty} \frac{(-1)^n}{\ln(n+2)} \cdot$$ The results of my algorithms are in agreement with the Pari/GP sumalt value of $0.92429989722293885595957$. But Wolfram Alpha gives the following approximated sum when entering

sum (-1)^n/(ln(n+2))

(a direct link from Math.SE will be mangled and does not work, here is the eq.):



  • Are the values from Pari and my algorithms correct?
  • Is there a closed form analytical result?
  • $\begingroup$ Concerning pari/gp's function sumalt (which will often return finite result even for divergent series) a trick is to gradually increase precision ( \p 100 and more) : the correct digits should remain as precision increases. $\endgroup$ Commented Dec 15, 2013 at 9:01
  • $\begingroup$ @Manu: Thank you for the info. I tested it on Raspberry Pi with a[n_] := 1/Log[n + 2]; EulerTransformation[0, 20] and can reproduce your result. But the result is always displayed with 4 digits even for error $\approx 10^{-8}$ and for nmax_ >= 22 I get obscure errors. $\endgroup$ Commented Jun 26, 2017 at 16:22
  • $\begingroup$ @TeM: This is equivalent to Pari/GP's sumalt. $\endgroup$ Commented Jun 27, 2017 at 10:25

3 Answers 3


For $x \in (0,1)$ consider

$$f(x) = \sum_{n=2}^{\infty} \frac{(-1)^n}{\log{(n+2)}} x^{\log{(n+2)}}$$


$$f'(x) = \frac{1}{x} \left [ 1-\left(1-2^{1+\log{x}}\right) \zeta(-\log{x})\right]$$

where $\zeta$ is the Riemann zeta function. Using $f(0)=0$, I get that the sum may be expressed in terms of the following integral:

$$\int_0^{\infty} du \left [ 1-\left ( 1 - \frac{1}{2^{u-1}}\right ) \zeta(u)\right]$$

  • $\begingroup$ Nice and tricky! But I guess your lower summation limit is 0? $\endgroup$ Commented Aug 26, 2013 at 13:23
  • $\begingroup$ @gammatester: I added and subtracted the $n=1$ term in the sum to get the zeta function; I hope I answered your question. $\endgroup$
    – Ron Gordon
    Commented Aug 26, 2013 at 13:24
  • $\begingroup$ Thank you for the final integral. I evaluated it with Pari and got the 0.92429989722293885595957 again. I will accept your answer because of the analytical result (and up-vote Artin's, and I hope I will avoid multipart questions in the future). $\endgroup$ Commented Aug 26, 2013 at 14:01
  • $\begingroup$ @gammatester: thanks. I like the integral representation because the series converges incredibly slowly. $\endgroup$
    – Ron Gordon
    Commented Aug 26, 2013 at 14:37

Since $\frac{1}{\log 2}-\frac 1{2\log 3}<1$ already, there is no way $1.07\dots$ can be anywhere close. To get high precision by hand, use Euler-Maclaurin. With Wolfram Alpha, I got $1.01845-2.45012\cdot 10^{-13}i$ . Remember, however, that WA was designed to impress calculus students and educators, not for any minimally non-trivial task...


some values.. enter image description here

with z=1 agree with your calculation pole in z=1 and the rest analitic continuation


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