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I try to find the sum of this alternating series : $\displaystyle \sum_{n=2}^{\infty} \frac{(-1)^{n}}{\ln(n)}$.

Since $\lim_{n\rightarrow \infty}\frac{1}{\ln(n)}=0$ and $\{\frac{1}{\ln(n)}\}$ is a decreasing sequence then by Alternating Series Test we know $\sum_{n=2}^{\infty} \frac{(-1)^{n}}{\ln(n)}$ is convergent.

Any hint or idea about how to find a closed form formula for the sum of this series ?

Jean Marie
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henry
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  • thanks,can you explain more about what the the closed for this series? – henry Sep 05 '16 at 22:23
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    I don’t believe it has a closed form. This is just speculation on my part and I don’t have a good reason to back this up though except this: if it had a closed form then I think it would be a famous result since the summand is such a nice and simple function. I've been around the block when it comes to sums and integrals and I have not heard about it. I can make this claim a bit more objective by computing the sum and using the inverse symbolic calculator. It finds no matches. – Winther Sep 05 '16 at 22:26
  • Mathematica 11 can't do it. –  Sep 05 '16 at 22:28
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    btw what you can do is to rewrite the sum as an integral: $$\sum_{n=2}^\infty \frac{(-1)^n}{\log(n)} = \int_0^\infty [1-\left(1-2^{1-x}\right) \zeta (x)]{\rm d}x$$ where $\zeta(s)$ is the Riemann zeta function. – Winther Sep 05 '16 at 22:46
  • $(\ln \ln x)' = \frac{1}{x \ln x}$, and $\sum_{n=2}^\infty (\ln \ln n) (-1)^n n^{-s}$ is maybe less hard – reuns Sep 06 '16 at 01:08
  • @Winther: how did you get that result? Sure looks tricky. – Ron Gordon Apr 11 '18 at 17:42
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    @RonGordon btw you should know this :) – Winther Apr 12 '18 at 13:09

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