# Does $\sum_{n=2}^{\infty} \frac{(-1)^n}{\ln(n)}$ have a closed form?

I was considering the sum $$\sum_{n=2}^{\infty} \frac{(-1)^n}{\ln(n)}$$ which I realizes converges extremely slowly (due to the slow growing nature of the logarithm). I was curious if someone knew a closed form for this in terms of literally ANY OTHER functions? Plugging this into Wolfram alpha didn't reveal anything particularly interesting:

https://www.wolframalpha.com/input/?i=sum+%28-1%29%5En%2F%28ln%28n%29%29+n+%3D2+to+infinity

I suppose that a good place to start would be try to understand the function $$\sum_{n=2}^{\infty} \frac{x^n}{\ln(n)}$$

And then evaluate it at $$x=-1$$. This function also was unfamiliar to me, so I tried to look at the even simpler object:

$$\sum_{n=1}^{\infty} \ln(n)x^n$$

Which does happen to have a restatement as a lambert series: See here: https://en.wikipedia.org/wiki/Lambert_series#Examples (particularly the section called Von Mangoldt Function). But even that series seems to be dependent on properties of prime numbers (perhaps there might be a way to get the riemann zeta function involved).

## Related:

Generally speaking $$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^q}$$ for small rational q (ex: $$q = \frac{1}{2}$$ or $$q = \frac{1}{3}$$) could be resolved using the riemann zeta function. The next slowest growing function that I could think of beyond polynomial roots was a logarithm which motivates this question.

Consider the series

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

Then

$$f'(x) = \sum_{n=1}^\infty (-1)^n n^x + 1 = 1 - \eta(-x) = 1 + (2^{1+x}-1)\zeta(-x)$$

where $$\eta$$ and $$\zeta$$ are the Dirchlet eta and Riemann zeta functions, respectively. Then the summation is given by

$$f(0) = \int_{-\infty}^0 1 - \eta(-x)\:dx = \int_0^\infty 1 - \eta(x)\:dx \approx0.9243$$

Use the polyogarithm $$\operatorname{Li}_s(z) = \sum_{k=1}^\infty\frac{z^k}{k^s},\qquad |z|<1 .$$ Then $$\operatorname{Li}_s(z) - z = \sum_{k=2}^\infty\frac{z^k}{k^s} \\ \int_0^\infty \big(\operatorname{Li}_s(z) - z\big)\;ds = \sum_{k=2}^\infty\frac{z^k}{\log(k)}$$
Compare this to $$\frac{\partial^n}{\partial s^n} \operatorname{Li}_s(z) = \sum_{k=1}^\infty \frac{z^k(\log k)^n}{k^s} \\ \frac{\partial^n}{\partial s^n} \operatorname{Li}_s(z)\Big\vert_{s=0} = \sum_{k=1}^\infty z^k(\log k)^n$$