\(\sum_{n=2}^{\infty} \frac{(-1)^n}{n\log{n}}\)
- 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.
Note: I don't have a solution for this.
Grigorios Kostakos
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 ... !
$$\int_1^\infty \left( 1+\left(2^{1-s}-1\right)\zeta(s) \right) \, \mathrm{d}s$$
However, the $\zeta$ function does not behave well under integrals. So, I would not expect a closed form to exist ... !
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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