Use mathematical induction to prove Σ n,k=1 (1/k(k+1)) = (n/n+1) for all n in Natural numbers?

Question

This is how far I can get:

p(n): nΣk=1 (1/k(k+1)) = (n/n+1) p(1): 1Σk=1 (1/(1+1)) = (1/1+1) => 1/2 = 1/2 p(1) is true.

Assume that p(k) is true. p(k) = kΣk=1, (1/k(k+1)) = k/k+1

Show that p(k+1) is true. p(k+1): k+1 Σ k=1, (1/k+1((k+1)+1)) = (k+1/(k+1)+1) => 1/(k+1)(k+2) = (k+1)/(k+2)

If this is correct, I am not sure how to finish from here. How can I simplify p(k+1) using the induction hypothesis p(k) to show that p(k+1) is also true.

Thanks!

Answer 1

Assume that it holds for $n=k$, i.e. $$\sum_{i=1}^{k}\frac{1}{i(i+1)}=\frac{k}{k+1}.$$ Then, we have $$\begin{align}\sum_{i=1}^{k+1}\frac{1}{i(i+1)}&=\left(\sum_{i=1}^{k}\frac{1}{i(i+1)}\right)+\frac{1}{(k+1)((k+1)+1)}\\&=\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}\\&=\frac{k(k+2)+1}{(k+1)(k+2)}\\&=\frac{(k+1)^2}{(k+1)(k+2)}\\&=\frac{k+1}{(k+1)+1}.\end{align}$$ Hence, it holds for $n=k+1$.

Answer 2

Inductive step

$$\sum_{k=1}^{n+1}\frac1{k(k+1)}=\sum_{k=1}^{n}\frac1{k(k+1)}+\frac1{(n+1)(n+2)}=\frac n{n+1}+\frac1{(n+1)(n+2)}\\=\frac{n^2+2n+1}{(n+1)(n+2)}=\frac{n+1}{n+2}$$

Alternative proof

We can find this result by telescoping

$$\sum_{k=1}^{n+1}\frac1{k(k+1)}=\sum_{k=1}^{n+1}\frac1{k}-\frac1{k+1}=1-\frac1{n+2}=\frac{n+1}{n+2}$$