$\sum\limits_{n=1}^{3N} a_n=S_{3N}$ is convergent and $\lim\limits_{n\to \infty} a_n=0$ then $\sum\limits_{n=1}^\infty a_n$ is convergent. Let $\displaystyle\sum_{n=1}^\infty a_n$ be infinity sum, and if $\displaystyle\sum_{n=1}^{3N} a_n=S_{3N}$ is convergent and $\lim\limits_{n\to \infty} a_n=0$ then $\displaystyle\sum_{n=1}^\infty a_n$ is convergent.
I couldnot start it from a reasonable point, I thought that only $\displaystyle\sum_{n=1}^{3N} a_n=S_{3N}$ is enough because it consists the whole convergent idea... any hint will be appreciated.
 A: We have
$$S_{3N+1}=S_{3N}+a_{3N+1} $$
and
$$S_{3N+2}=S_{3N+1}+a_{3N+2} $$
thus
$$\lim_{N\to+\infty}S_{3N}=\lim_{N\to+\infty} S_{3N+1}=\lim_{N\to+\infty} S_{3N+2} .$$
this means that the sequence $(S_N) $ is convergent and so is the series $\sum a_n .$ 
A: Note that $S_{3N}, S_{3N+1}, S_{3N+2}$ all have the same limit (say $L$) since $S_{3N+1}=S_{3N}+a_{3N+1}$, $S_{3N+2}=S_{3N}+a_{3N+1}+a_{3N+2}$ and $a_n\to 0$ as $n\to \infty$.
Thus, there exists $n_0$ such that $N\geq n_0$ implies
$$
|S_{3N}-L|<\epsilon \quad \text{and}\quad  |S_{3N+1}-L|<\epsilon\quad {\text{and}}\quad |S_{3N+2}-L|<\epsilon
$$
 Since every positive integer is congruent modulo $3$ to $0$, $1$ or $2$, it follows that $S_{N}$ converges to $L$.
A: Sketch: Fix $\varepsilon>0$.


*

*Since $S_{3N}$ is convergent, let $L$ be its limit.  Then, there exists an $N_0$ so that if $N>N_0$, then $|S_{3N}-L|<\varepsilon$.

*Since $a_n$ converges to $0$, there exists an $n_0$ so that if $n>n_0$, $|a_n|<\varepsilon$.
Therefore, if $M>n_0$ and $M>3N_0$, then there are three cases depending on if $M\equiv 0\pmod 3$, $M\equiv 1\pmod 3$, and $M\equiv 2\pmod 3$.  The $M\equiv 0\pmod 3$ case is given.  When $M\equiv 2\pmod 3$, then $M=3K+2$, so 
$$
\left|\sum_{n=1}^Ma_n-L\right|=\left|\sum_{n=1}^{3K}a_n+a_{3K+1}+a_{3K+2}-L\right|
\leq\left|\sum_{n=1}^{3K}a_n-L\right|+\left|a_{3K+1}\right|+\left|a_{3K+2}\right|\leq 3\varepsilon.
$$
Can you take it from here?
