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

Calculus (Integrals, Series)
Grigorios Kostakos
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$\sum_{n=2}^{\infty} \frac{(-1)^n}{n\log{n}}$

Evaluate the convergent series $\mathop{\sum}\limits_{n=2}^{\infty} \frac{(-1)^n}{n\log{n}}\,.$

Note: I don't have a solution for this.
Grigorios Kostakos
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Re: $\sum_{n=2}^{\infty} \frac{(-1)^n}{n\log{n}}$

Basically it equals to

$$\int_1^\infty \left( 1+\left(2^{1-s}-1\right)\zeta(s) \right) \, \mathrm{d}s$$

However, the $\zeta$ function does not behave well under integrals. So, I would not expect a closed form to exist ... !
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$