Value of $\sum_{n=0}^{\infty} \frac{(-1)^n}{\ln(n+2)}$

Question

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.):

$$\sum_{n=0}^{\infty}\dfrac{(-1)^n}{\log(2+n)}\approx1.00766524110155\ldots$$

Questions:

• Are the values from Pari and my algorithms correct?
• Is there a closed form analytical result?

Answer 1

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

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

Then

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

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:

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

Answer 2

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...

Answer 3

some values..

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