Convergence of series proof Let $ \left(a_{n}\right)_{n=1}^{\infty} $ be monotonic decreasing of real positive numbers.
Let $ \left(n_{k}\right)_{k=1}^{\infty} $ be strictly increasing natural numbers sequence, such that exists $ M\in\mathbb{R} $ such that $ n_{k+1}-n_{k}\leq M\left(n_{k}-n_{k-1}\right) $ for any $ k \in \mathbb{R} $ .
Prove that $ \sum_{n=1}^{\infty}a_{n} $ converge, $ \iff $ $ \sum_{k=1}^{\infty}\left(n_{k+1}-n_{k}\right)a_{n_k} $ converge.
On one direction, I assumed $ \sum_{n=1}^{\infty}a_{n} $ converge, and tried to use the comparison test by limit. But the index confusing me.
I want to show that $ \lim\frac{a_{n}}{\left(n_{k+1}-n_{k}\right)a_{n_{k}}} $ exists, and maybe it will help me to say something smart about the convergence of $ \sum_{n=1}^{\infty}\left(n_{k+1}-n_{k}\right)a_{n_{k}} $ but not sure what would be the variable of the limit ?
Also, we need to conclude from this proof that $ \sum_{n=1}^{\infty}a_{n} $ converges $ \iff $ $ \sum_{k=1}^{\infty}k^{2}a_{k^{2}} $ converges.
I've thinking about those questions a while and couldnt proceed. Any ideas will help.
Thanks in advance
 A: Let's start with the observation that  from the decreasing monoticity of $(a_n)$ it follwos that all $(a_n)$ must be non-negative, as otherwise neither $a_n$ nor $(n_{k+1}-n_k)a_{n_k}$ could tend to zero, which is a necessary condition for the resoective sums to converge. In other words, if some $a_n$ is negative, all the following are as well (and smaller = bigger in absolute value), so both sums trivially diverge.
So let's assume
$$\forall n: a_n \ge 0 \tag{1} \label{pos}.$$
The key to solve the problem is to realize that when you consider $(n_{k+1}-n_k)a_{n_k} =\underbrace{a_{n_k} + a_{n_k} + \ldots + a_{n_k}}_{(n_{k+1}-n_k) \text{ terms}}$, that some partial sums of $\sum_{n=1}^{\infty} a_n$ and $\sum_{k=1}^{\infty} (n_{k+1}-n_k)a_{n_k}$ have actually the same number of summands, so can be compared well:
$$
\begin{eqnarray}
\sum_{n=1}^{\infty} a_n  =  a_1 + a_2 + \ldots +a_{n_1-1} & + & \underbrace {a_{n_1} + a_{n_1+1} + \ldots + a_{n_2-1}}_{n_2-n_1 \text{ terms}}  & +  & \underbrace {a_{n_2} + a_{n_2+1} + \ldots + a_{n_3-1}}_{n_3-n_2 \text{ terms}} & + & \ldots\\
\sum_{k=1}^{\infty} (n_{k+1}-n_k)a_{n_k}  =  && \underbrace{a_{n_1} + a_{n_1} + \ldots + a_{n_1}}_{n_2-n_1 \text{ terms}} & +  & \underbrace {a_{n_2} + a_{n_2} + \ldots + a_{n_2}}_{n_3-n_2 \text{ terms}} & + & \ldots
\end{eqnarray}
$$
The first $n_1-1$ summands of $\sum_{n=1}^{\infty} a_n$ have no corresponding terms in $\sum_{k=1}^{\infty} (n_{k+1}-n_k)a_{n_k}$, but a fixed number of initial summands doesn't change convergence of a series. Afterwards, each $n_{k+1}-n_k$ summands of $\sum_{n=1}^{\infty} a_n$ correspond to one term $(n_{k+1}-n_k)a_{n_k}$ or, when expanded as above, $n_{k+1}-n_k$ summands of $a_{n_k}$.
That leads to the inequality I hinted at in my comment for any integer $s \ge 1$:
$$\sum_{k=n_1}^{n_s-1} a_k \le \sum_{k=1}^{s-1}(n_{k+1}-n_k)a_{n_k} \tag2 \label{eqrueck}$$
because the right hand sum replaces each block $\underbrace {a_{n_k} + a_{n_k+1} + \ldots + a_{n_{k+1}-1}}_{n_{k+1}-n_k \text{ terms}}$ from the left hand sum with a block $\underbrace{a_{n_k} + a_{n_k} + \ldots + a_{n_k}}_{n_{k+1}-n_k \text{ terms}}$, which is at least as big. The latter follows from the $(a_n)$ sequence being monotonically decreasing.
Since all involved summands are positive, because of \eqref{pos}, the boundedness of the partial sums is equivalent for the sums to converge. So from \eqref{eqrueck} the $\Longleftarrow$ of the equivalence follows immediately, as the right hand side of \eqref{eqrueck} is then bounded by finite $\sum_{k=1}^{\infty} (n_{k+1}-n_k)a_{n_k}$, so arbitrarily long partial sums of $\sum_{n=1}^{\infty} a_n$ are bounded.
To prove the other direction of the equivalence, we use the same "block comparison" method, but we obviously need to do it in the other direction, replacing $\underbrace {a_{n_k} + a_{n_k+1} + \ldots + a_{n_{k+1}-1}}_{n_{k+1}-n_k \text{ terms}}$ with something that is less or equal. Replacing  each summand with $a_{n_{k+1}}$ will make the sum smaller or equal, again based on $(a_n)$ being monotonically decrasing, so we have
$$\underbrace {a_{n_k} + a_{n_k+1} + \ldots + a_{n_{k+1}-1}}_{n_{k+1}-n_k \text{ terms}} \ge  (n_{k+1}-n_k)a_{n_{k+1}} \ge \frac1M(n_{k+2}-n_{k+1}) a_{n_{k+1}}, \tag3 \label{eqhin}$$
with the last inequality being a slightly rearranged and index-shifted version of the $M$-condition.
So if we assume that that $\sum_{n=1}^{\infty} a_n$ converges, then from \eqref{eqhin}
follows
$$\sum_{k=n_1}^{n_s-1} a_k \ge \sum_{k=1}^{s-1}(n_{k+2}-n_{k+1})a_{n_{k+1}} =  \sum_{k=2}^{s}(n_{k+1}-n_{k})a_{n_{k}}, $$
so the right hand side (which is a partial sum of $\sum_{k=1}^{\infty}(n_{k+1}-n_{k})a_{n_{k}}$, missing only the first term) is bounded, so converges.
