# The sum $\sum_{n=1}^\infty1/(aN-b)^k$ for the case $k>2$

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,

Greg

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.