Monotone non-increasing functions (sequence) relations Not sure what is the topic of this question, but it seems like real analysis.

The question itself: Given $a(q) : \mathbb{N} \rightarrow \mathbb{R}$ monotone non-increasing positive sequence, s.t. $\sum_{q=1}^{\infty}a(q) = \infty$, there exists a monotone non-increasing sequence $a'(q)$, s.t. also - $\sum_{q=1}^{\infty}a'(q) = \infty$ and for all $t \geq 1$, $$\frac{a'(q)}{a(tq)} \xrightarrow{q \rightarrow \infty} 0$$

Any clues?
First, notice that, for each fixed $t$, the series $\sum_{q=1}^\infty a(tq)$ diverges.  The reason is that, if this series converged, then so would the series $\sum_{q=1}^\infty a(tq+1)$, $\sum_{q=1}^\infty a(tq+2)$, $\dots,\sum_{q=1}^\infty a(tq+t-1)$ because, by the "non-increasing" assumption, they're dominated by $\sum_{q=1}^\infty a(tq)$.  But the sum of all these convergent series would be all of $\sum_{q=1}^\infty a(q)$ except for the first $t-1$ terms, and this is assumed to diverge.

Second, notice that, still for fixed $t$, $\sum_{q=1}^\infty a(tq)/t$ diverges; we've just multiplied the divergent series from the preceding paragraph by a constant $1/t$.

With these observations in mind, we can define the required $a'(q)$ sequence as follows.

First, since $\sum_{q=1}^\infty a(q)$ diverges, find an $N(1)$ so large that $\sum_{q=1}^{N(1)}a(q)\geq1$, and set $a'(q)=a(q)$ for $1\leq q\leq N(1)$.  This ensures that, no matter how we define $a'(q)$ for larger $q$ (as long as it's positive), $\sum_{q=1}^\infty a'(q)$ will be at least $1$.  Call this stage $1$ of the definition.

Next, find $N(2)$ so large that $\sum_{q=N(1)+1}^{N(2)}a(2q)/2\geq1$.  Such an $N(2)$ exists because, as shown above, $\sum_{q=1}^\infty a(2q)/2$ diverges.  Set $a'(q)=a(2q)/2$ for $N(1)&lt;q\leq N(2)$.  You thereby ensure that $\sum_{q=1}^\infty a'(q)$ will be at least $2$.  Call this stage $2$.

Continue similarly, setting $a'(q)=a(3q)/3$ at stage $3$ for $N(2)&lt;q\leq N(3)$, where $N(3)$ is chosen so large that these $a'(q)$'s add up to at least $1$ and therefore $\sum_{q=1}^\infty a'(q)$ will be at least $3$.  And so forth for all later stages.

Since you keep making $\sum_{q=1}^\infty a'(q)$ at least $1$ bigger at each step, you clearly get this series to diverge.  Furthermore, if you fix some $t$ and then look at what happens at stages $s$ from $t$ on, you're setting $a'(q)=a(sq)/s\leq a(tq)/s$, so you have $a'(q)/a(tq)\leq 1/s$.  Since this holds for all stages $s\geq t$, and since $1/s\to0$ as $s\to\infty$, you get $a'(q)/a(tq)\to 0$ as required.
If $a(q)$ is a constant sequence, put $a'(q)=q.a(q)$ and you're done.

If $a(q)$ is decreasing, the subsequence $a(tq)$ is also decreasing.

If $\lim\limits_{q\to\infty}a(q)=-\infty$, take $a'(q)=1$ and you're done ($\sum\limits_{q=1}^{+\infty}a'(q)=\infty$).

If $\lim\limits_{q\to\infty}a(q)=\ell\in\mathbb{R}^*$, let $a'(q)=\frac{1}{q}$ so $\lim\limits_{q\to \infty}a'(q)=0$ and $\sum\limits_{q=1}^{+\infty}a'(q)=\infty$

If $\lim\limits_{q\to\infty}a(q)=0$, take $a'(q)=0$