$\sum_{n=2}^{\infty} (-1)^n \frac{1}{\log n}$ =? The alternating series, $s_n=\sum_{i=2}^{n} (-1)^i \frac{1}{\ln i}$, (with $\ln$ being the natural logarithm) converges for $n\to\infty$ which can be seen e.g. by the Leibniz test. Can the limit be expressed in a finite closed expression of otherwise known constants? 
I got interested while thinking about the values of the sine integral at integer multiples of $\pi$ where I suspect some relation. 
 A: Doing my usual
pairing of even and odd,
if
$s_n
=\sum_{i=2}^{n} (-1)^i \dfrac{1}{\ln i}
$
then
$\begin{array}\\
s_{2n+1}
&=\sum_{i=2}^{2n+1} (-1)^i \dfrac{1}{\ln i}\\
&=\sum_{i=1}^{n} (\dfrac{1}{\ln (2i)}-\dfrac{1}{\ln (2i+1)})\\
&=\sum_{i=1}^{n} \dfrac{\ln (2i+1)-\ln (2i)}{\ln (2i)\ln (2i+1)}\\
&=\sum_{i=1}^{n} \dfrac{\ln (1+1/2i)}{\ln (2i)\ln (2i+1)}\\
&<\sum_{i=1}^{n} \dfrac{1/2i}{\ln (2i)\ln (2i+1)}\\
&=\sum_{i=1}^{n} \dfrac{1}{2i\ln (2i)\ln (2i+1)}\\
\end{array}
$
and this converges by comparison with
$\sum \dfrac1{i \ln^2 i}
$.
However,
this does converge
quite slowly.
The integral test,
using
$\int_2^n \dfrac{dt}{t\ln^2(t)}
=\dfrac1{\ln(2)}-\dfrac1{\ln(n)}
$
shows how slowly.
This, of course,
agrees with the result that
the sum of an alternating series
is between any two
consecutive partial sums.
For example,
if $n = 10^{10}$,
the error is about
$\dfrac1{2\ln(10^{10})}
=\dfrac1{20\ln(10)}
\gt .01
$.
If we apply
Euler's transform
(https://en.wikipedia.org/wiki/Series_acceleration),
$\begin{array}\\
s
&=\sum_{n=0}^{\infty} (-1)^n \dfrac{1}{\ln (n+2)}\\
&=\sum_{n=0}^{\infty} \dfrac{(-1)^n}{2^{n+1}}\sum_{k=0}^n(-1)^k\binom{n}{k} \dfrac{1}{\ln (n-k+2)}\\
&\approx\sum_{n=0}^{\infty} \dfrac{(-1)^n}{2^{n+1}}(\dfrac1{\ln(x+2)})^{(n)}|_{x=2}\\
&\approx\sum_{n=0}^{\infty} \dfrac{(-1)^n}{2^{n+1}}((-1)^{n-1}\dfrac{(n-1)!}{(x+2)^n\ln(x+2)})|_{x=2}\\
&=-\sum_{n=0}^{\infty} \dfrac{(n-1)!}{2^{3n+1}\ln(4)}\\
\end{array}
$
and this actually diverges!
(modulo any errors on my part)
That's enough for now.
A: Another comment, with no closed form at the end.
Start with the polylogarithm
$$
\sum_{n=1}^\infty a^n n^{-s} = \mathrm{Li}_s(a)
$$
So that
$$
\sum_{n=1}^\infty (-1)^n n^{-s} = \mathrm{Li}_s(-1)
\\
\sum_{n=2}^\infty (-1)^n n^{-s} = 1 + \mathrm{Li}_s(-1)
$$
Integrate with respect to $s$ from $0$ to $+\infty$
$$
\sum_{n=2}^\infty (-1)^n \frac{1}{\log(n)}
= \int_0^\infty\big(1+\mathrm{Li}_s(-1)\big)ds \approx 0.92429989722293885595957018136
$$
Unfortunately, this also has no known closed form as far as I can find.
