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Is there a general formula for this sum? $$I(a,b,c)=\sum_{n=0}^\infty \frac{1}{(an+b)^c}, a> 0, b>0, c>1$$ I noticed that the first few a, b, and c values yielded seemingly different results, but all look related in some way to the zeta function. Here are a few values of I. $$I(2, 1, 2)= \frac{\pi^2}{8}$$ $$I(2, 1, 3)= \frac{7\zeta(3)}{8}$$ $$I(3, 1, 2)= \frac{\psi^{(1)}(\frac{1}{3})}{9}$$ $$I(1,2,2)= \frac{\pi^2}{6} - 1$$ $$I(\pi,\pi,\pi)= \frac{\zeta(\pi)}{\pi^{\pi}}$$ Don't pay too much attention to the bounds for a and b as I don't really know if anything changes should the values be negative. Perhaps if a general formula does not exist, maybe one does when fixing certain values such as setting b = 1?

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  • $\begingroup$ no, there isnt a closed formula for that $\endgroup$ – Masacroso Jan 8 '19 at 0:17
  • $\begingroup$ Some special values can be obtained through the zeta function, e.g. $a=1, b=0$. For example, for $n \in \Bbb N$ $$\zeta (2n)={\frac {(-1)^{n+1}B_{2n}(2\pi )^{2n}}{2(2n)!}}$$ $${\displaystyle \zeta (-n)=(-1)^{n}{\frac {B_{n+1}}{n+1}}}$$ Other identities can be found at en.wikipedia.org/wiki/Riemann_zeta_function. . $\endgroup$ – Eevee Trainer Jan 8 '19 at 0:20
  • $\begingroup$ But I imagine you already know the previous going by the tags and some of the values. To my knowledge there is no closed form known for odd integer inputs of the zeta function (maybe part of why Apery's constant, $\zeta(3)$, is not known to be transcendental or not). Which, in light of that, (a) might mean this question will likely not yield anything more than partial answers and (b) might be better off on Math Overflow, but that's just a guess. At the very least it would be a very nontrivial question and I'm not the one to ask if things are better on MO. $\endgroup$ – Eevee Trainer Jan 8 '19 at 0:20
  • $\begingroup$ @EeveeTrainer Thank you for your comment. I've edited the question to be more open and reasonable. Further, I'm not too knowledgable about the zeta function given my high school calculus knowledge. Any information would help me! $\endgroup$ – Suchetan Dontha Jan 8 '19 at 0:22
  • $\begingroup$ With respect to one particular case ($a=b=1, c=2$), there is a discussion on it in this video - youtube.com/watch?v=aYc9ManKC-s - I haven't watched it yet so I don't have anything to offer there. Just figured you might find it useful. $\endgroup$ – Eevee Trainer Jan 8 '19 at 0:25
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$I(a,b,c) =\sum_{n=0}^\infty \frac{1}{(an+b)^c} $

Some simple cases.

$I(1,1,c) =\sum_{n=0}^\infty \frac{1}{(n+1)^c} =\sum_{n=1}^\infty \frac{1}{n^c} =\zeta(c) $

$I(2,2,c) =\sum_{n=0}^\infty \frac{1}{(2n+2)^c} =2^{-c}\sum_{n=1}^\infty \frac{1}{n^c} =2^{-c}\zeta(c) $

$\begin{array}\\ I(2,1,c) &=\sum_{n=0}^\infty \frac{1}{(2n+1)^c}\\ &=\sum_{n=0}^\infty \frac{1}{(2n+1)^c}+\sum_{n=0}^\infty \frac{1}{(2n+2)^c}-\sum_{n=0}^\infty \frac{1}{(2n+2)^c}\\ &=\sum_{n=1}^\infty \frac{1}{n^c}-I(2, 2, c)\\ &=I(1, 1, c)-I(2, 2, c)\\ &=\zeta(c)-2^{-c}\zeta(c)\\ &=(1-2^{-c})\zeta(c)\\ \end{array} $

If $b$ is an integer,

$I(1,b,c) =\sum_{n=0}^\infty \frac{1}{(n+b)^c} =\sum_{n=b}^\infty \frac{1}{n^c} =\sum_{n=1}^\infty \frac{1}{n^c}-\sum_{n=1}^{b-1} \frac{1}{n^c} =\zeta(c)-\sum_{n=1}^{b-1} \frac{1}{n^c} $

Note that $I(a,b,c) =\sum_{n=0}^\infty \frac{1}{(an+b)^c} =a^{-c}\sum_{n=0}^\infty \frac{1}{(n+b/a)^c} =a^{-c}\zeta(c, b/a) $ the Hurwitz zeta function.

See https://en.wikipedia.org/wiki/Hurwitz_zeta_function for a discussion of this.

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  • $\begingroup$ Thanks for your answer! This was exactly what I was looking for when I first thought of the question. I can't wait to study this function more. $\endgroup$ – Suchetan Dontha Jan 8 '19 at 0:34
  • $\begingroup$ See my addition at the end of my answer. $\endgroup$ – marty cohen Jan 8 '19 at 0:43
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While not a complete answer, if $c$ is a positive integer greater than 1 the sum can be written in terms of the polygamma function $\psi^{(m)}(z)$.

From the series representation for the polygamma function, namely $$\psi^{(m)}(z) = (-1)^{m + 1} \Gamma (m + 1) \sum_{n = 0}^\infty \frac{1}{(n + z)^{m + 1}}, \quad m > 0, z \notin 0,\mathbb{Z}^-$$ if we rewrite your sum as $$I(a,b,c) = \frac{1}{a^c} \sum_{n = 0}^\infty \frac{1}{(n + b/a)^{(c - 1) + 1}},$$ in terms of the polygamma function the sum becomes $$I(a,b,c) = \frac{(-1)^c}{a^c \Gamma (c)} \psi^{(c - 1)} \left (\frac{b}{a} \right ).$$ Here $c > 1$ where $c \in \mathbb{N}$, $a > 0$, and $b \neq 0,-a,-2a,\ldots$

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For $a=b=c=k$, where k is a constant, it seems as though $I(k, k, k) = \frac{\zeta(k)}{k^k}$

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