Let $\left ( a\_{n} \right )\_{n=1}^{\infty }$ be a sequence of real numbers and $s\_{k}=a\_{1}+a\_{2}+\cdot \cdot \cdot +a\_{k}$ be the $k$-th
partial sum.

Suppose that $\lim\_{}a\_{n}=0$, and there exists a $m\in \mathbb{N}$ such that the sequence $\left ( s\_{mk} \right )\_{k=1}^{\infty }=(s\_{m},s\_{2m},s\_{3m}...)$ converges.

Prove that $\sum a\_{n}$ converges.

Here, I have some expert's solution, which I doubt if it's really correct.

Proof: $\forall \varepsilon >0$,

since $\lim\_{}a\_{n}=0$, $\exists N\_{1}>0$ s.t $\left | a\_{n} \right |<\frac{\varepsilon }{4m} $ $\forall n>N\_{1}$.

Also, since $\left ( s\_{mk} \right )\_{k=1}^{\infty }$ converge, $\exists N\_{2}>0$ s.t $\left |a\_{km+1}+\cdot \cdot \cdot +a\_{lm}\right |<\frac{\varepsilon }{2}$ $\forall l>k>N\_{2}.
$

Let $N=\max \{ N\_{1},m(N\_{2}+2) \}$.

Then, $\forall q\geq p>N$,$$|a\_{p}+a\_{p+1}+\cdot \cdot \cdot+a\_{q}|$$ $$\leq|a\_{km+1}+\cdot \cdot \cdot+a\_{lm}|+2m\cdot\frac{\varepsilon }{4m}$$ $$<\varepsilon$$ where $k$ is the least integer s.t $km+1\geq p$
, and $l$ is the largest integer s.t $lm\leq q$. $$\blacksquare $$

The most tedious part to understand is "how do we know there is some number which is $0$ (mod $m$), and some other number which is $1$ (mod $m$)?" (What if $p=q$?)

And, "why do we need to set $N$ as this solution did?"

Lastly, "how do we know that there are less than $m$ terms between $p$ and $km+1$, and there are less than $m$ terms between $lm$ and $q$?"