<s>[INST] Proof of convergence of $s_n = \frac{1}{2\cdot 1} + \frac{1}{3\cdot 2} + \cdots + \frac{1}{n(n+1)}$ I am really bad at this, but I am trying my best to understand.
If I am given a sequence of partial sums such that

$$s_n = \frac{1}{2\cdot1} + \frac{1}{3\cdot2} + \cdots + \frac{1}{n(n+1)}$$
Prove that $s_n \to 1$.

I always have trouble tackling this. I know I have to find an $N$ s.t. $|s_n - 1 | < \epsilon$. But to what extent can I go to select this $N$? Any tips on selecting $N$ or is it just intuition based?
Or perhaps could I do the following:
I know
$$\begin{align}
s_1 & = \dfrac{1}{2} \\
s_2 & = \dfrac{2}{3} \\
s_3 & = \dfrac{3}{4} \\
& \vdots \\
s_n & = \dfrac{n}{n+1}
\end{align}
$$
and it is trivial to show $\left|\dfrac{n}{n+1} - 1\right|$ converges by doing the following:
$$\left|\frac{n}{n+1} - 1\right| = \left|\frac{n}{n+1} - \frac{n+1}{n+1}\right| = \left|\frac{1}{n+1}\right| \le \frac{1}{N+1} < \epsilon$$
But I am not sure if I have to prove why $s_n = \dfrac{n}{n+1}$. If so I think I would have to do so by induction? Do I have to show how I got $\dfrac{n}{n+1}$ formally and rigorously or can I assume my audience follows?
 [/INST] Can you show by induction that
$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots+\frac{1}{n(n+1)}=\frac{n}{n+1}. $$
Indeed, for $n=1$ is true because
$$\frac{1}{1\cdot 2}=\frac{1}{2}=\frac{1}{1+1}. $$
Now, suppose that the equality is true for $n=k$, i.e.,
$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots+\frac{1}{k(k+1)}=\frac{k}{k+1}, $$
then, adding $\frac{1}{(k+1)(k+2)}$ we have
$$ \frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots+\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}=\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}=$$
$$=\frac{k(k+2)+1}{(k+1)(k+2)}=\frac{(k^2+2k+1)}{(k+1)(k+2)}=\frac{k+1}{k+2}=\frac{k+1}{(k+1)+1},$$
then the equality follow for induction. I have helped this part.
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