Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

From my work on the Goldbach conjecture I have formulated the following series $ζ[ς]=\sum\limits_{f=2}^∞ \frac{1}{f^ς}-\frac{2}{(f+1)f-2}$ where $ς$ a natural number. If $ς=2$ we have the series


Does anyone know for which $ς$ these series are convergent?

share|cite|improve this question
up vote 2 down vote accepted

Your sum can be rewritten as

$$ \sum_{f=2}^{\infty} \left( \frac{1}{f^s} - \frac{2}{(f+1)f-2} \right) = \sum_{f=2}^{\infty} \frac{1}{f^s} - \sum_{g=2}^{\infty} \frac{2}{(g+1)g-2}. $$

Now, the right-hand sum can be evaluated explicitly since it telescopes. Indeed,

$$ \begin{align*} \sum_{g=2}^{\infty} \frac{2}{(g+1)g-2} &= 2\sum_{g=2}^{\infty} \frac{1}{(g+1)g-2} \\ &= 2\sum_{g=2}^{\infty} \frac{1}{(g-1)(g+2)} \\ &= \frac{2}{3} \sum_{g=2}^{\infty} \left(\frac{1}{g-1} - \frac{1}{g+2}\right) \\ &= \frac{2}{3} \left(1 \color{red}{- \frac{1}{4}} + \frac{1}{2} \color{blue}{- \frac{1}{5}} + \frac{1}{3} \color{violet}{- \frac{1}{6}} \color{red}{+ \frac{1}{4}} \color{violet}{- \frac{1}{7}} \color{blue}{+ \frac{1}{5}} \color{violet}{- \frac{1}{8}} + \cdots\right) \\ &= \frac{2}{3}\left(1 + \frac{1}{2} + \frac{1}{3}\right) \\ &= \frac{11}{9}, \end{align*} $$

where the terms in $\color{red}{\text{red}}$ cancel with each other, the terms in $\color{blue}{\text{blue}}$ cancel with each other, and the terms in $\color{violet}{\text{violet}}$ cancel with something that isn't shown.

Then, since

$$ \sum_{f=2}^{\infty} \frac{1}{f^s} = \sum_{f=1}^{\infty} \frac{1}{f^s} - 1 = \zeta(s) - 1, $$

where $\zeta$ is the Riemann zeta function, we see that your sum is equal to

$$ \sum_{f=2}^{\infty} \left( \frac{1}{f^s} - \frac{2}{(f+1)f-2} \right) = \zeta(s) - 1 - \frac{11}{9} = \zeta(s) - \frac{20}{9}. $$

The Riemann zeta function converges for all $s$ with $\Re(s) > 1$, so the same is true for your sum.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.