Evaluate $\sum_{k=2}^\infty{\frac{(-1)^k}{\ln k}}$

Question

I'm wondering if there is an exact solution of $$S=\sum_{k=2}^\infty{\frac{(-1)^k}{\ln k}}$$ I let $$S(x)=\sum_{k=2}^\infty{\frac{(-1)^kk^x}{\ln k}}$$ So$$S'(x)=\sum_{k=2}^\infty{(-1)^kk^x}=1+(2^{x+1}-1)\zeta(-x)$$ and $S(-\infty)=0$
Then I can get$$S=S(0)=\int_{-\infty}^0{1+(2^{x+1}-1)\zeta(-x)dx}$$ I have no idea about the integral.

Answer 1

The given series is convergent by Leibniz' test, but in order to compute an accurate numerical approximation of it we need to accelerate the convergence of its partial sums, since $\log k$ diverges quite slowly. A first simple manipulation is $$ \sum_{k\geq 2}\frac{(-1)^k}{\log k} = \sum_{n\geq 1}\left(\frac{1}{\log(2n)}-\frac{1}{\log(2n+1)}\right)=\sum_{n\geq 1}\frac{\log\left(1+\frac{1}{2n}\right)}{\log(2n)\log(2n+1)} $$ where the main term of the RHS now behaves like $\frac{1}{2n\log^2(2n)}$. Still unsatisfactory, but this is approximately $\frac{1}{2}\left(\frac{1}{\log(2n)}-\frac{1}{\log(2n+2)}\right)$, so by creative telescoping $$ \sum_{k\geq 2}\frac{(-1)^k}{\log k}=\frac{1}{2\log 2}+\frac{1}{2}\sum_{n\geq 1}\left[\frac{1}{\log(2n)}-\frac{2}{\log(2n+1)}+\frac{1}{\log(2n+2)}\right] $$ where the main term of the last series behaves like $\frac{1}{4n^2\log^2(n)}$. The next step is to approximate $\frac{1}{2}\left[\frac{1}{\log(2n)}-\frac{2}{\log(2n+1)}+\frac{1}{\log(2n+2)}\right]$ with $\frac{1}{4}\left[\frac{1}{2n\log^2(2n)}-\frac{1}{(2n+2)^2\log^2(2n+2)}\right]$ to get $$\sum_{k\geq 2}\frac{(-1)^k}{\log k}=\frac{1}{2\log 2}+\frac{1}{8\log^2(2)}+\\+\frac{1}{4}\sum_{n\geq 1}\small{\left[\frac{2}{\log(2n)}-\frac{4}{\log(2n+1)}+\frac{2}{\log(2n+2)}-\frac{1}{2n\log^2(2n)}+\frac{1}{(2n+2)\log^2(2n+2)}\right]}.$$ The main term of the last series behaves like $\frac{1}{n^4\log^2(n)}$. Each term is negative but the very first term is already $\geq -\frac{1}{4}$, hence

$$ S-\left(\frac{1}{2\log 2}+\frac{1}{8\log^2(2)}\right) \in \left(-\frac{1}{10},0\right)$$ and $S\approx 0.9243$.

Answer 2

$$\sum _{k=0}^{\infty } \frac{z^k}{\log (k+2)}$$ have a zero near to the Z=0,662

fo more values have a pole in z=1