Generating: $\frac{1}{2\ln2}-\frac{1}{3\ln3}+\frac{1}{4\ln4}...$

Question

Help the King out. Brand new stuff to him. I know (-1)^n has something do to with it but I don't know what else.

Write in sigma notation:

$\frac{1}{2\ln2}-\frac{1}{3\ln3}+\frac{1}{4\ln4}...$

Answer 1

Let's start with $1$ minus the Dirichlet eta function : $$1- \eta(s)=\sum_{n=2}^\infty\frac{(-1)^n}{n^s}=\frac 1{2^s}-\frac 1{3^s}+\cdots$$ Since $\,\displaystyle \int \frac 1{n^s} ds=-\frac 1{n^s\log(n)}\;$ we deduce $\,\displaystyle \frac 1{n\log(n)}=\int_1^\infty \frac 1{n^s} ds\;$ and :

$$\sum_{n=2}^\infty\frac{(-1)^n}{n\,\log(n)}=\sum_{n=2}^\infty(-1)^n\int_1^\infty\frac{1}{n^s}\,ds=\int_1^\infty 1-\eta(s)\; ds$$

You may rewrite $\eta(s)$ as $\,\eta(s)=\left(1-2^{1-s}\right)\zeta(s)\,$ but this shouldn't make the last integral much easier. I don't know a closed form for this last integral (integrals over $\zeta$ don't usually give closed forms as opposed to the corresponding series).

Hoping this helped anyway,