Question

For all $n\ge 1$, Prove that:
$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots +\frac{1}{n(n+1)}=\frac{n}{n+1}$

Answer 1

Step $\left(1\right):$ Assume given statement
Let the given statement be $P\left(n\right)$, i.e.
$P\left(n\right):\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots +\frac{1}{n(n+1)}=\frac{n}{n+1}$

Step $2:$ Checking statement $P\left(n\right)$ for $n=1$
Put $n=1$ in $P\left(n\right)$, we get
$P\left(1\right):\frac{1}{1\cdot 2}=\frac{1}{1+1}$
$\Rightarrow \frac{1}{2}=\frac{1}{2}$, which is true
Thus $P\left(n\right)$ is true for $n=1$.

Step $3:$ $P\left(n\right)$ for $n=K$
Put $n=K$ in $P\left(n\right)$, and assume that $P\left(K\right)$ is true for some natural number $K$ i.e.
$P\left(K\right):\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots +\frac{1}{K(K+1)}=\frac{K}{(K+1)}\cdots \left(1\right)$

Step $\left(4\right):$ Checking statement $P\left(n\right)$ for $n=K+1$
Now, we need to prove that $P(K+1)$ is true whenever $P\left(K\right)$ is true, we have
$P(K+1):\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots +\frac{1}{K(K+1)}+\frac{1}{(K+1)(K+2)}$
$=(\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\cdots +\frac{1}{K(K+1)})+\frac{1}{(K+1)(K+2)}$
$=\frac{K}{K+1}+\frac{1}{(K+1)(K+2)}$ {from eq. $\left(1\right)$}
$=\frac{K(K+2)+1}{(K+1)(K+2)}$
$=\frac{(K^2+2K+1)}{(K+1)(K+2)}$
$=\frac{(K+1{)}^2}{(K+1)(K+2)}$
$=\frac{K+1}{K+2}$
We can write it as, $\frac{K+1}{(K+1)+1}$
Thus, $P(K+1)$ is true whenever $P\left(K\right)$ is true.
Final Answer:
Hence, by principle of Mathematical Induction, $P\left(n\right)$ is true for all natural numbers.