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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

$\frac{1}{2^2}-\frac{1}{2}+\frac{1}{3^2}-\frac{1}{5}+\frac{1}{4^2}-\frac{1}{9}+\frac{1}{5^2}-\frac{1}{14}$....

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

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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.

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