How to use mathematical induction to verify: $\sum_{i=1}^{n}\frac{1}{i(i+1)} = \frac{n}{n+1}$

Question

How to use mathematical induction to verify: $\sum_{i=1}^{n}\frac{1}{i(i+1)} = \frac{n}{n+1}$

I have already tried it myself: see here

but it is just not working out...

Thanks in advance!

Answer 1

Base Step:- If $n=1$, then

$$\sum_{i=1}^{1}\frac{1}{i(i+1)} = \frac{1}{1(1+1)}=\frac12$$ and $$\frac{n}{n+1}=\frac{1}{1+1}=\frac12$$ Since $n=k$,the base step is true.

Inductive Hypothesis:- Assume $P(k)$ is true. $$\sum_{i=1}^{k}\frac{1}{i(i+1)} = \frac{k}{(k+1)}$$

Now show that $P(k+1)$ is true $$\sum_{i=1}^{k+1}\frac{1}{i(i+1)} = \frac{k+1}{k+1+1}=\frac{k+1}{k+2}$$

Now, $$\sum_{i=1}^{k+1}\frac{1}{i(i+1)} =\sum_{i=k}^{k}\frac{1}{i(i+1)}+\frac{1}{k+1(k+2)}$$ $$=\frac{k}{k+1}+\frac{1}{k+1(k+2)}$$ $$=\frac{k^2+2k+1}{(k+1)(k+2)}$$ $$=\frac{(k+1)^2}{(k+1)(k+2)}$$ $$=\frac{k+1}{k+2}$$ Therefore, $p(k+1)$ is true.

By the principle of mathematical induction $\sum_{i=1}^{n}\frac{1}{i(i+1)} = \frac{n}{n+1}$

Answer 2

For n=1: $$ \begin{align} \sum_{i=1}^{1}\frac{1}{i\left(i+1\right)}&=\frac{1}{1(1+1)}=\frac{1}{2}\\ \frac{n}{n+1}&=\frac{1}{1+1}=\frac{1}{2} \end{align} $$ hence proved for n=1.

Assume it holds true for n=k, where k is any natural number. Therefore: $$ \begin{align} \sum_{i=1}^{k}\frac{1}{i\left(i+1\right)}&=\frac{k}{k+1}\\ \end{align} $$ Hence inducing that for $k=k+1$: $$ \begin{align} \sum_{i=1}^{k+1}\frac{1}{i\left(i+1\right)}&=\sum_{i=1}^{k}\frac{1}{i\left(i+1\right)}+\frac{1}{(k+1)(k+2)}\\ &=\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}\\ &=\frac{k(k+2)+1}{(k+1)(k+2)}\\ &=\frac{k^2+2k+1}{(k+1)((k+1)+1)}\\ &=\frac{(k+1)^2}{(k+1)((k+1)+1)}\\ &=\frac{(k+1)}{(k+1)+1} \end{align} $$ and then write a conclusion :)