<s>[INST] Proof by induction for "sum-of" Prove that for all $n \ge 1$:
$$\sum_{k=1}^n \frac{1}{k(k+1)} = \frac{n}{n+1}$$
What I have done currently:
Proved that theorem holds for the base case where n=1.
Then:
Assume that $P(n)$ is true. Now to prove that $P(n+1)$ is true:
$$\sum_{k=1}^{n+1} \frac{1}{k(k+1)} = \frac{n}{n+1} + n+1$$
So:
$$\sum_{k=1}^{n+1} \frac{1}{k(k+1)} = \frac{n+1}{n+2}$$
However, how do I proceed from here?
 [/INST] For n=1:
$$
\sum_{k=1}^{1} 1/(k(k+1))=1/2=1/(1+1)
$$
Let's prove it for $n+1$:
$$
\sum_{k=1}^{n+1} \frac{1}{k(k+1)}= \sum_{k=1}^{n}+\frac{1}{(n+1)(n+2)}\\
$$
But using our assumption for $P(n)$ :
$$
\sum_{k=1}^{n} \frac{1}{k(k+1)}=\frac{n}{n+1}
$$
So the sum will be:
$$
\sum_{k=1}^{n+1} \frac{1}{k(k+1)}=\frac{n}{(n+1)} + \frac{1}{(n+1)(n+2)}\\
=\frac{n(n+2)+1}{(n+1)(n+2)}\\
=\frac{n^2+2n+1}{(n+1)(n+2)}\\
=\frac{{(n+1)}^{2}}{(n+1)(n+2)}\\
=\frac{n+1}{n+2}
$$
Hence proved.
Note that you made a mistake in your last part where you wrote:
$$
\sum_{k=1}^{n+1} \frac{1}{k(k+1)}= \frac{n}{n+1}+n+1
$$
</s>