I got stuck trying to calculate the following sum: $$\sum_{k=1}^{\infty} \frac{1}{k^\alpha}\left(\frac{1}{k}-\frac{1}{k+1}\right)$$ where $\alpha > -1$ (this ensures that the sum converges). Any help would be greatly appreciated!

  • $\begingroup$ It's the same as $\sum\limits_{j=0}^\infty (-1)^j \zeta(\alpha+2+j)$, but does it help? $\endgroup$ – user90369 Dec 11 '17 at 20:13
  • $\begingroup$ Thanks for the observation, but I do not know how to use it to get the end result... $\endgroup$ – Metod Jazbec Dec 11 '17 at 21:20
  • $\begingroup$ @MetodJazbec: what is the end result? Such series does not appear to have a nice closed form for any $\alpha>-1$. $\endgroup$ – Jack D'Aurizio Dec 11 '17 at 22:50

I'm on the bus so I can't reason well. In any case the series reads also

$$\sum_{k = 1}^{+\infty} \frac{1}{k^{\alpha}}\frac{1}{k(k+1)}$$

Which you may majorize just to have an idea of its behaviour:

$$\sum_{k = 1}^{+\infty} \frac{1}{k^{\alpha}}\frac{1}{k(k+1)} \leq \sum_{k = 1}^{+\infty} \frac{1}{k^{\alpha}}\frac{1}{k^2} = \zeta(2+\alpha)$$

Where $\zeta(p)$ is the Riemann Zeta Function.

  • $\begingroup$ Courtesy of Jack D'Aurizio's PDFs, we know the closely related series:$$\sum_{k=1}^\infty\frac1{k(k+1)^n}=n-\sum_{k=2}^n\zeta(k)$$See page 7. $\endgroup$ – Simply Beautiful Art Dec 13 '17 at 2:13
  • $\begingroup$ @SimplyBeautifulArt What a cute pdf, although there is nothing "superior". Elementary maths one learns in Analysis 1. Thanks for the input anyway! $\endgroup$ – Von Neumann Dec 13 '17 at 16:08

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