On the sums $\sum_{i=1}^{n} \sum_{j=0}^{m-1} \dfrac{a_j}{mi-j}$ as $n \to \infty$ 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).
 A: 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 $\sum_{k=0}^\infty \left(\frac{1}{5k+1} - \frac{1}{5k+2} - \frac{1}{5k+3} + \frac{1}{5k+4} \right)$
