<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] This step is wrong:
$$\sum_{k=1}^{n+1} 1/k(k+1) = n/(n+1) + n+1$$
it should read:
$$\sum_{k=1}^{n+1} \dfrac{1}{k(k+1)} = \dfrac{n}{n+1} + \dfrac{1}{(n+1)(n+2)}$$
so we have:
$$\sum_{k=1}^{n+1} \dfrac{1}{k(k+1)} = \dfrac{n(n+2)+1}{(n+1)(n+2)}$$
$$= \dfrac{n^2+2n+1}{(n+1)(n+2)}$$
$$= \dfrac{(n+1)^2}{(n+1)(n+2)}$$
$$= \dfrac{n+1}{n+2}$$
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