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Question: How much is $$\sum_{N=1}^\infty \frac{1}{aN-b}$$ where $a$ is a positive integer, and $b\geq -1$ is an integer?

Could it be a polygamma function or a Lerch Transcendent? I'm not sure, especially as I'm subtracting $b$, not adding it. (Learned about these particular functions a few days ago, as a result of having had an earlier question answered.)

Motivation: A figurate number $$p(s,N) = \frac{1}{2}N\Bigl( (s-2)N-(s-4)\Bigr).$$ I want to compute the value of the sequence $[1/p(s,N)]^k$, where $k$ is a positive integer and the sum goes from $N=1$ to $N=\infty$. The computational challenge is to figure out the case for $k=1$ and then, for higher $k$, do a binomial expansion of $$\frac{2}{s-4}\sum_{N=1}^{\infty}\left(\frac{s-2}{(s-2)N - (s-4)} - \frac{1}{N}\right)^k$$

In doing the expansion, it's easy to arrange all of the middle terms (excluding the first and last) so that they have a denominator that is $(N)[(s-2)N - (s-4)]$ raised to some power, which permits the results for these terms to be expressed in terms of lower k's. The last term, $\sum\left(\frac{(-1}{N}\right)^k$is just the appropriate zeta function. Its the first term, $$\sum \left(\frac{s-2}{(s-2)N-(s-4)}\right)^k$$that needs to be cracked in order to have an explicit expression.

I've evaluated this expression (for all $k$) for $s=3$ (Triangular numbers) and $s=6$ (Hexagonal numbers), the simplest cases where figurate numbers are concerned.

With an answer to the question above, I'll just take $a=(s-2)$ and $b=(s-4)$, and multiply the result by the numerator $(s-2)$. Thanks,


P.S. Am happy to share the explicit results I've found so far for the triangular and hexagonal cases for those who are interested.

My bad! My question title should have been $\sum_{n=1}^\infty1/(aN-b)^k$ for the case $k>2$. I fear my motivation discussion didn't make that sufficiently clear. (For $k=1$, you have $1/(aN-b) - 1/N$, which does converge for the figurate numbers, even though either term taken individually is divergent.)

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I can't speak to your motivation, but I think you are a looking at a minor variation of the harmonic series. Any sum of the form $\sum_{N=1}^\infty \frac{1}{aN - b}$ is divergent. – Adam Saltz Mar 9 '12 at 17:10

The series diverges. For sufficiently large $N$ it consists of positive terms; using the limit comparison test with $\sum\frac{1}{n}$ we get $$\lim_{n\to\infty}\frac{1/n}{1/(an-b)} = \lim_{n\to\infty}\frac{an-b}{n} = a.$$ Since $0\lt a\lt \infty$ and $\sum\frac{1}{n}$ diverges, both series diverge.

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