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

Calculus (Integrals, Series)
Post Reply
User avatar
Grigorios Kostakos
Founder
Founder
Posts: 461
Joined: Mon Nov 09, 2015 1:36 am
Location: Ioannina, Greece

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

#1

Post by Grigorios Kostakos »

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
User avatar
Riemann
Posts: 176
Joined: Sat Nov 14, 2015 6:32 am
Location: Melbourne, Australia

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

#2

Post by Riemann »

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 ... !
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
Post Reply

Create an account or sign in to join the discussion

You need to be a member in order to post a reply

Create an account

Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute

Register

Sign in

Who is online

Users browsing this forum: No registered users and 21 guests