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