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Suppose $f(n)$ is a periodic function with period $q$. Now from this paper we get that if $\displaystyle\sum_{n=1}^{q}f(n)=0$ then $\displaystyle\sum_{n=1}^{\infty}\frac{f(n)}{n}=-\frac{1}{q}\displaystyle\sum_{a=1}^{q}f(a)\psi(\frac{a}{q})$, where $\psi$ is digamma function.

My question is whether we can find a generalization of this or not? I mean:

$f(n)$ is a periodic function with period $q$, but $\displaystyle\sum_{n=1}^{q}f(n)$ may not be zero. In that case for $\Re(s)>1$ define $F(s)=\displaystyle\sum_{n=1}^{\infty}\frac{f(n)}{n^s}$. Of course we can write $F(s)$ as finite linear combination of special values of Hurwitz Zeta function. Apart from that, is there any way to write $F(s)$ as a finite sum (e.g. finite linear combination of special values of digamma function like previous case)?

Any suggestion or reference will be appreciated.

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2 Answers 2

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See theorem 1 in S. Adhikari, N. Saradha, T.N. Shorey, R. Tijdeman Transcendental infinite sums Indag. Math. (N.S.), 12 (1) (2001), pp. 1–14 (seems to be available for free online, if not see here.)

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  • $\begingroup$ Thanks Professor Rivin. I have seen this theorem before. But pardon my ignorance, I don't understand how theorem 1 relates to my problem! I was asking about $\Re(s)>1$. Say, what is the sum $\sum_{n=1}^{\infty}\frac{\{\frac{n}{k}\}}{n^s}$? By the result mentioned in my question, sum is $(\log k)^{-1}$ if $s=1$. I wonder what about $s=2$, can we find a closed form? $\endgroup$ Jul 16, 2013 at 9:07
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Thanks to everyone whoever thought over this problem. I have asked Professor Murty (one of the authors of the paper mentioned in the question) about this question. He told me that, of course such generalization exists. In fact, those same authors had another paper which encounters this generalization problem.

It is very clear that according to the question, $L(s,f):=\displaystyle\sum_{n=1}^{\infty}\frac{f(n)}{n^s}$ is analytic in $\mathbb{C}$ except $s=1$ where the L-function has a simple pole with residue $\frac{1}{q}\displaystyle\sum_{n=1}^qf(n)$.

For $1<k \in \mathbb{N}$, we can have $L(k,f)=\frac{(-1)^k}{(k-1)!q^k}\displaystyle\sum_{a=1}^qf(a)\psi_{k-1}(\frac{a}{q})$, where $\psi_k$ is Polygamma function. For details of the proof one may look at this paper .

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