Series, limits and convergence. 
Theorem $\,\bf3.3.1.\;$ If the series $$\sum_{n=1}^\infty a_n$$ is convergent then $\lim\limits_{n\to\infty}a_n=0$.
  Proof. Let $s_n=\sum_{k=1}^n a_k.$ Then by the definition the limit $\lim_n s_n$ exists. Denote is by $s.$ Then of course $\lim_n s_{n-1}=s$. Note that $a_n=s_n-s_{n-1}$ for $n\geqslant2$. Hence $$\lim_{n\to\infty}a_n=\lim_{n\to\infty}s_n-\lim_{n\to\infty}s_{n-1}=s-s=0.\qquad\square$$

Can somebody please help me to explain this proof? I do not understand why we define $s_n$ like this and why it follows $\lim_n s_{n-1} = s$, further why is $a_n=s_n-s_{n-1}$?
Thanks
 A: 
I do not understand why we define $s_n$ like this.

Check the definition of $\sum \limits_{n=1}^\infty (a_n)$.

Why it follows $\lim \limits _ns_{n−1}=s$?

In general, given a sequence $(x_n)_{n\in \mathbb N}$, one has $\lim \limits_{n\to \infty}\left((x_n)_{n\in \mathbb N}\right)=\lim \limits_{n\to \infty}\left((x_{n-k})_{n\in \mathbb N_{\large n\ge k}}\right)$, for all $k\in \mathbb N$, in particular for $k=1$. This is easy to prove by definition.

Why $a_n=s_n-s_{n-1}$?

Let $n\in \mathbb N$ and assume $n\ge 2$. One has $$s_n-s_{n-1}=\sum \limits_{k=1}^n(a_k)-\sum \limits_{k=1}^{n-1}(a_k)=\left(a_n+\sum \limits_{k=1}^{n-1}(a_k)\right)-\sum \limits_{k=1}^{n-1}(a_k).$$
A: When we want to see if something converges, one way we can do it is define something as the partial sum, we did that with $s_n$ in this case.
$a_n = s_n - s_{n-1}$ follows since $s_n$ is the sum of $a_m$ with $m$ from 1 to $n$, so $s_n = a_1 + a_2 + ... + a_n$ and $s_{n-1} = a_1 + a_2 + ... + a_{n-2} + a_{n-1}$.
This means that when you do $s_n - s_{n-1}$ you're actually doing this:
$$a_1 - a_1 + a_2 - a_2 + ... + a_{n-2} - a_{n-2}  + a_{n-1} - a_{n-1} + a_n$$
Notice how $a_n$ is the only term not subtracted from itself.
They have defined $s = \lim_{n\rightarrow\infty}s_n$ (which they wrote as $\lim_{n}s_n$ without the arrow to infinity, but I think it makes more sense to write the arrow and the infinity)
Then they say it follows that if $s = \lim_{n\rightarrow\infty}s_n$ then $s = \lim_{n\rightarrow\infty}s_{n-1}$. This is justified since $s_{n-1}$ is really just one term behind from $s_n$ and at the limit that one that it's behind dosen't matter anymore.
