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Timeline for answer to Dirichlet series and Riemann zeta function by user10676

Current License: CC BY-SA 3.0

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Jul 30, 2014 at 13:34 comment added Snufsan Nevermind, Got it!
Jul 30, 2014 at 13:21 comment added Snufsan Still i cant figure how we got the $2k+1$ factor. we have $$\frac{1}{(1-p^{-s})^2} = \sum_{n=0}^{\infty} \frac{d(p^k)}{p^{ks}} $$ and then what?
Jul 30, 2014 at 13:15 comment added Semiclassical @Snufsan: Note that the denominator is the square of an infinite series, and so has a tidy expression as a sum. Multiplying by the numerator then adds to that series a 'shifted' copy of itself, and that gives the simple form shown.
Jul 30, 2014 at 12:58 comment added Snufsan Could you please explain the transition from $$\frac{1+p^{-s}}{(1-p^{-s})^2}$$ to the infinite sum?
Jul 30, 2014 at 12:53 vote accept Snufsan
Jul 30, 2014 at 12:57
Jul 30, 2014 at 12:50 history answered user10676 CC BY-SA 3.0