I was working on my proof by induction and I'm second guessing myself on solving the left hand side.
Prove: $$ \frac{1}{1\times 2}+\frac{1}{2 \times 3}+ \frac{1}{3 \times 4}+ \ldots + \frac{1}{n\times(n+1) }= \frac{n}{n+1} \qquad \forall n \in \mathbb{N} $$
Proof: Principle of Mathmatical Induction (P.M.I.)
Assume: P(n):1/(1*2)+1/(2*3)+1/(3*4)+⋯+1/(n*(n+1) )=n/(n+1),∀n∈N
First we need to show that P(1) is true.
P(1):1/((1)*((1)+1) )=((1))/((1)+1)
P(1):1/2=1/2
∴ P(1) is true.
Next,assume P(k) is true for some k∈N.
P(k): 1/(1*2)+1/(2*3)+1/(3*4)+⋯+1/(k*(k+1) )=k/(k+1),is true.
Show P(k+1) is true.
P(k+1): 1/(1*2)+1/(2*3)+1/(3*4)+⋯+1/(k*(k+2) )=k/(k+2),for some k∈N
L.H.S.= 1/(1*2)+1/(2*3)+1/(3*4)+⋯+1/(k*(k+2) )
= [1/(1*2)+1/(2*3)+1/(3*4)+⋯+1/(k*(k+1) )]+1/(k*(k+2) )
= k/(k+1)+1/(k*(k+2) )
= k/(k+1)+1/(k^2+2k)
=0/1+1/(k^2+2k)
= k/(k+2)
= R.H.S.
∴ P(k+1) is true.
Hence,by P.M.I.
P(n) is true for all n∈N.∎
For $P(k+1)$ the denominator should be $(k+1)(k+2)$, not $k(k+1)$, so we need to show that $$\frac{1}{1\times 2}+\frac{1}{2 \times 3}+ \frac{1}{3 \times 4}+ \ldots + \frac1{k\times(k+1)}+\frac{1}{(k+1)\times(k+2) }= \frac{k+1}{k+2}$$ where $n$ is replaced by $k+1$. From the induction hypothesis, it is sufficient to show that \begin{align}\frac k{k+1}+\frac{1}{(k+1)\times(k+2) }= \frac{k+1}{k+2}&\impliedby k(k+2)+1=\frac{k+1}{k+2}\times(k+1)(k+2)\\&\impliedby k^2+2k+1=(k+1)^2\end{align} which is true as $k>0$.