Exercise 2.5.13 in Introduction to Real Analysis by Jiri Lebl 
Let $s_k$ be the $k$th partial sum of $\sum x_n$.
a) Suppose that there exists a $m \in \mathbb{N}$ such that $\lim_{k\to\infty} s_{mk}$ exists and $\lim x_n = 0$. Show that $\sum x_n$ converges.
b) Find an example where $\lim_{k \to \infty} s_{2k}$ exists and $\lim x_n \not=0$ (and therefore $\sum x_n$ diverges).
c) (Challenging) Find an example where $\lim x_n =0$, and there exists a subsequence $\{s_{k_j}\}$ such that $\lim_{j \to \infty} s_{k_j}$ exists, but $\sum x_n$ still diverges.

I know that the series converges iff $s_k$ converges. So, I think that the question a) simply follows from this equivalence (am I right?). For b), I think that this series should be the one that is alternating and each term cancel out only when the series is even partial sums, but I can't find such series. I have no idea for c).
This question is a bit challenging to me. I appreciate if you give some help.
 A: For $a):$ Given $\epsilon >0,$ take $k_1\in \Bbb N$ such that $k_1< k\le k'\implies |s_{km}-s_{k'm}|<\epsilon/3.$ Take $k_2\in \Bbb N$ with $k_2\ge k_1$ and such that $k_2m< j\implies |x_j|<\epsilon/(3m).$
Now for $k_2m\le n<n',$ consider $k,k'\in \Bbb N$ where $km>n\ge (k-1)m$ and  $k'm>n'\ge (k'-1)m.$
We have $k_1< k\le k'$ so $$|s_{km}-s_{k'm}|<\epsilon/3.$$ We have  $k_2m< n+1 \le km $ so $$|s_n-s_{km}|=|\sum_{j=n+1}^{km}x_j\,|\le \sum_{j=n+1}^{km}|x_j|<$$ $$<(km-n)\cdot \epsilon/(3m)\le m\cdot \epsilon/(3m)=\epsilon/3.$$ Similarly we have $$|s_{k'm}-s_{n'}|<\epsilon /3.$$
So $|s_n-s_{n'}|\le |s_n-s_{km}|+|s_{km}-s_{k'm}|+|s_{k'm}-s_{n'}|<\epsilon.$
For $c):$. Take a sequence $(b_n)_{n\in \Bbb N}$ of members of $(0,1]$ such that $\lim_{n\to \infty}b_n=0$ and $\sum_{n\in \Bbb N}b_n=\infty.$ E.g. $b_n=1/n.$
Let $a_1=b_1.$
If $s_n\ge 1$ then $a_{n+1}=-b_{n+1}.$
If $s_n\le 0$ then $a_{n+1}=b_{n+1}.$
If $s_n\in (0,1)$ and $a_n>0$ then $a_{n+1}=b_{n+1}.$ 
If $s_n\in (0,1)$ and $a_n<0$ then $a_{n+1}=-b_{n+1}.$
For any $r\in [0,1],$ the sequence $S=(s_n)_{n\in \Bbb N}$ has a subsequence converging to $r.$ For your Q, it suffices to show that $S$ has a subsequence converging to $1$ and another subsequence converging to $0.$
$S$ is like a snail that wanders back and forth from approximately $1$ to approximately $0$ in "eventually smaller" steps.
