Prove that $\sum a_{n}$ converges. (I have some doubts on an expert's solution) 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 $q$?"
 A: I’ll start with your last question. The choice of $k$ ensures that
$$(k-1)m+1<p\le km+1\,,\tag{1}$$
so
$$(km+1)-p<(km+1)-\big((k-1)m+1\big)=m\,.$$
Similarly, the choice of $\ell$ ensures that $\ell m\le q<(\ell+1)m$, so
$$q-\ell m<(\ell+1)m-\ell m=m\,.$$
For your first question, if $\ell\le k$, the set $\{i\in\Bbb N:km+1\le i\le\ell m\}$ is empty, and the term $|a_{km+1}+\ldots+a_{\ell m}|$ is $0$; there are fewer than $2m$ terms altogether, and we have
$$|a_p+\ldots+a_q|\le 2m\cdot\frac{\epsilon}{4m}=\frac{\epsilon}2\,.$$
Finally, we need $N\ge N_1$ to ensure that $|a_n|<\frac{\epsilon}{4m}$ for each $n>N$; we use that when we conclude that the terms $|a_i|$ with $p\le i\le km$ or $\ell m<i\le q$ contribute less than $\frac{\epsilon}2$.
We also need to ensure that $n>N$ implies that
$$|a_{km+1}+\ldots+a_{\ell m}|<\frac{\epsilon}2\,.\tag{2}$$
We know that $(2)$ holds if $\ell>k>N_2$. From $(1)$ we can infer that $k=\left\lceil\frac{p-1}m\right\rceil$, so if $p>m(N_2+1)$, then $p-1>mN_2$ (since $m\ge 1)$, and
$$k=\left\lceil\frac{p-1}m\right\rceil\ge N_2+1>N_2\,.$$
Similarly, the inequalities $\ell m\le q<(\ell+1)m$ imply that $\ell=\left\lfloor\frac{q}m\right\rfloor$, so that if $q>m(N_2+1)$, then
$$\ell=\left\lfloor\frac{q}m\right\rfloor\ge N_2+1>N_2\,.$$
Thus, making sure that $N\ge m(N_2+1)$ ensures that $(2)$ holds whenever $n>N$. It appears that we could get away with letting $N=\max\{N_1,m(N_2+1)\}$.
A: The proof is indeed correct although I dare say it could be presented in a clearer and better articulated form. Allow me to attempt to deliver such a presentation, in a more general setting to which the claim above naturally extends. Consider a real Banach space $\left(E, +, \cdot, \lVert \bullet \rVert\right)$ together with a sequence $x \in E^{\mathbb{N}}$ such that $\displaystyle\lim_{n \to \infty}x_n=0_E$ and such that there exists a fixed $r \in \mathbb{N}^{\times}=\mathbb{N}\setminus\{0\}$ for which the sequence $t=\left(s_{rn}\right)_{n \in \mathbb{N}}$ is convergent, where $s=\left(\displaystyle\sum_{k=0}^nx_k\right)_{n \in \mathbb{N}}$ represents the sequence of partial sums of $x$.
As a detail of notation, for any $x \in \mathbb{R}$ let $[x]\colon=\max\{n \in \mathbb{Z}\mid n \leqslant x\}$ denote the integer part of $x$. For arbitrary natural numbers $m, n \in \mathbb{N}$ let us also write $[m, n]_{\mathbb{N}}=[m, n] \cap \mathbb{N}$. For any set $M$ its cardinality will be denoted as $|M|$. Recall that in general $\left|[m, n]_{\mathbb{N}}\right|=n-m+1$ for any $m, n \in \mathbb{N}$ such that $m \leqslant n+1$.
We claim that the series associated to $x$ converges, in other words that the sequence $s$ is convergent. In order to prove this we employ the completeness of the norm on our space $E$ and establish the fact that $s$ is a Cauchy sequence. Consider an arbitrary $\rho>0$. Since $x_n \xrightarrow{n \to \infty} 0_E$ we gather the existence of a number $k \in \mathbb{N}$ such that $\lVert x_n \rVert < \frac{\rho}{4r}$ for any $n \geqslant k$.
Furthermore, as sequence $t$ converges it is Cauchy and therefore there exists $h \in \mathbb{N}$ such that $\lVert t_m-t_n \rVert <\frac{\rho}{2}$ for any numbers $m, n \geqslant h$. Bearing in mind that for $m \geqslant n$ we have the relation $t_n-t_m=\displaystyle\sum_{k=rm+1}^{rn}x_k$, the condition involving $h$ can be expressed as saying that $\left\lVert \displaystyle\sum_{k=rm+1}^{rn} x_k\right\rVert <\frac{\rho}{2}$ for any numbers $m, n$ such that $h \leqslant m \leqslant n$.
We shall now show that for any numbers $m, n \in \mathbb{N}$ such that $m, n \geqslant \max\{k+r-1, rh\}$ the relation $\lVert s_n-s_m \rVert < \rho$ holds, thus achieving our goal. Without any loss of generality assume that $m \leqslant n$. As $r \neq0$, by virtue of the fundamental theorem of division with remainder there exist unique numbers $m', n', p, q \in \mathbb{N}$ such that $m=m'r+p$, $n=n'r+q$ and $p, q<r$. Since $m \leqslant n$ it follows with necessity that $m' \leqslant n'$ (otherwise we would have $n'<m'$ entailing $n'+1 \leqslant m'$ and hence $n=n'r+q<\left(n'+1\right)r\leqslant m'r \leqslant m'r+p=m$, which constitutes a contradiction). We also remark that $m'=\left[\frac{m}{r}\right]$ and since $m \geqslant rh$ we infer $h \leqslant \frac{m}{r}$ and thus $h \leqslant m'$ by definition of the integer part.
We proceed to exhibit the partition:
$$\left[m'r+1, \left(n'+1\right)r\right]_{\mathbb{N}}=\left[m'r+1, m'r+p\right]_{\mathbb{N}} \cup [m+1, n]_{\mathbb{N}} \cup \left[n'r+q+1, \left(n'+1\right)r\right]_{\mathbb{N}},$$
on the grounds of which we can write:
$$\displaystyle\sum_{l=m'r+1}^{\left(n'+1\right)r}x_l=\displaystyle\sum_{l=m'r+1}^{m'r+p}x_l+\displaystyle\sum_{l=m+1}^n x_l+\displaystyle\sum_{l=n'r+q+1}^{\left(n'+1\right)r}x_l$$
and hence:
$$s_n-s_m=\sum_{l=m+1}^n x_l=\sum_{l=m'r+1}^{\left(n'+1\right)r}x_l-\left(\sum_{l=m'r+1}^{m'r+p}x_l+\sum_{l=n'r+q+1}^{\left(n'+1\right)r}x_l\right)=t_{n'+1}-t_{m'}-\left(\sum_{l=m'r+1}^{m'r+p}x_l+\sum_{l=n'r+q+1}^{\left(n'+1\right)r}x_l\right).$$
Let us remark that since $p \leqslant r-1$ and $m 
\geqslant k+r-1$ we have $m'r=m-p \geqslant m+1-r \geqslant k$ and also $h \leqslant m' \leqslant n' < n'+1$, which means - on the basis of the upper bounds made explicit in the previous paragraphs - we can therefore infer the following estimates:
\begin{align*}
\lVert s_n-s_m \rVert &\leqslant \lVert t_{n'+1}-t_{m'} \rVert + \left\lVert \sum_{l=m'r+1}^{m'r+p}x_l \right\rVert + \left\lVert \sum_{l=n'r+q+1}^{\left(n'+1\right)r}x_l \right\rVert \\
&\leqslant \lVert t_{n'+1}-t_{m'} \rVert + \sum_{l=m'r+1}^{m'r+p} \lVert x_l \rVert + \sum_{l=n'r+q+1}^{\left(n'+1\right)r} \lVert x_l \rVert \\
&<\frac{\rho}{2}+\sum_{l=m'r+1}^{m'r+p} \frac{\rho}{4r}+\sum_{l=n'r+q+1}^{\left(n'+1\right)r} \frac{\rho}{4r} \\
&=\frac{\rho}{2}+\frac{\rho}{4r}\left(\left|\left[m'r+1, m'r+p\right]_{\mathbb{N}} \right|+\left| \left[n'r+q+1, \left(n'+1\right)r \right]_{\mathbb{N}} \right|\right) \\
&=\frac{\rho}{2}+\frac{\rho}{4r}\left(p+r-q\right)\\
&<\frac{\rho}{2}+2r \frac{\rho}{4r}\\
&=\rho,
\end{align*}
where for the final inequality we have taken into account the relations $p<r$ and $r-q \leqslant r$.
This concludes the proof.
