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This was inspired by Evaluating $\sum_{k=0}^\infty \left(\frac{1}{5k+1} - \frac{1}{5k+2} - \frac{1}{5k+3} + \frac{1}{5k+4} \right)$

Let $s(n) =\sum_{i=1}^{n} \sum_{j=0}^{m-1} \dfrac{a_j}{mi-j} $ and $A = \sum_{j=0}^{m-1} a_j $.

(1) Show that $\big|s(n)-\dfrac{A\ln n}{m}\big| $ is bounded as $n \to \infty$ so that $\lim_{n \to \infty} s(n)$ exists if and only if $A = 0$.

(2) Show that $\lim_{n \to \infty} \big|s(n)-\dfrac{A\ln n}{m}\big| $ exists.

(3) Give a closed form for $\lim_{n \to \infty} \big|s(n)-\dfrac{A\ln n}{m}\big| $.

I have shown (1) but not (2) or (3).

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  • $\begingroup$ Just out of curiosity : may I ask what have been the steps for $(1)$ ? The problem is nice. Cheers. $\endgroup$ – Claude Leibovici Dec 13 '17 at 11:26
  • $\begingroup$ I'll post what I have in a day or so. $\endgroup$ – marty cohen Dec 13 '17 at 11:34
  • $\begingroup$ (2) can be reduced to (1), because replacing $s(n)$ by $s(n)-\dfrac{A\,H_n}{m}$ is equivalent to replacing $a_0$ by $a_0-A$, and $H_n-\ln n\to\gamma$. (3) can be done in terms of the values $\psi_0(j/m)$, and those can be expressed with the help of Gauss's Digamma Theorem. $\endgroup$ – Professor Vector Dec 13 '17 at 13:58
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I try to respond to all three issues at once. I admit that my method might lack some rigour.

Define

$$s=\sum _{i=1}^n \left(\sum _{j=0}^{m-1} \frac{a_j}{i\; m-j}\right)$$

$$A=\sum _{j=0}^{m-1} a_j$$

$$B=\sum _{j=0}^{m-1} j \;a_j$$

Expanding the summand of the s-sum for $i\to \infty$ up to the order $\frac{1}{i^2}$ we have

$$\frac{a_j}{i\; m-j}=\frac{1}{i\; m}\frac{a_j}{1-\frac{j}{i\; m}} \simeq \frac{a_j}{i\; m}+ \frac{j\; a_j}{(i\; m)^2}$$

Doing the j-sum gives for $s$

$$s\to \sum _{i=1}^{\infty } \left(\frac{A}{i\, m}+\frac{B}{(i\, m)^2}\right)$$

The sum over $i$ of the first term is obviously divergent unless $A = 0$.

EDIT 14.12.17

In order to determine the sum in a general form we use contour integration similar to http://algo.inria.fr/flajolet/Publications/FlSa98.pdf.

We find, using the harmonic number $H(-i)$ as the kernel to produce the series starting at $i=1$, that $s$ is the negative sum of the residues at the poles of the summand giving

$$s=- \sum _{j=1}^{m-1} \frac{a_j}{m} H(-\frac{j}{m})\tag{1}$$

Notice that $a_0$ does not appear because of the condition $A=0$.

I have checked (1) with the example of Evaluating $\sum_{k=0}^\infty \left(\frac{1}{5k+1} - \frac{1}{5k+2} - \frac{1}{5k+3} + \frac{1}{5k+4} \right)$

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  • $\begingroup$ That's essentially my method, so I'll accept it. $\endgroup$ – marty cohen Dec 13 '17 at 19:58
  • $\begingroup$ I have corrected my "quick shot" on the result for the sum in an edit. $\endgroup$ – Dr. Wolfgang Hintze Dec 14 '17 at 7:14

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