Use mathematical induction to prove that $\sum_{j=1}^n \frac{1}{(j^2-1)}$ = $\frac{3}{4}$ - $\frac{2n+1}{2n(n+1}$ for all n $\geq$ 2

Question

so I know how to do induction and all I'm just struggling with the algebra for proving the case for n = k+1. My proof is below:

For the case n=2 we have

$\sum_{j=1}^2 \frac{1}{j^2-1}$ = $\frac{1}{3}$ and $\frac{3}{4}$ - $\frac{2*2 + 1}{2n(n+1)}$ = $\frac{1}{3}$.

Thus the equality holds for n = 2.

Assume for n = k for k $\in$ $\mathbb{Z}$ that the equality holds.

For the case n = k + 1 we must show that

$\sum_{j=1}^{k+1} \frac{1}{j^2-1}$ = $\frac{3}{4}$ - $\frac{2(k+1)+1}{2(k+1)((k+1)+1)}$

For the LHS we have:

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

I'm not gonna type all the algebra out because that's what I'm having trouble with and there's just a lot of it.

For the RHS we have:

$\frac{3}{4}$ - $\frac{2(k+1)+1}{2(k+1)((k+1)+1)}$ = $\frac{3}{4}$ - $\frac{2(k+1)+1}{(2k+2)(k+2)}$ = $\frac{3}{4}$ - $\frac{2k+3}{2k^2+6k+4}$

When you expand both sides I'm not seeing any way to manipulate them to equal one another... I know it's just basic algebra that's messing this up because the instructions are to prove it.

Answer 1

\begin{align} \frac34 - \frac{2k+1}{2k(k+1)} + \frac{1}{(k+1)^2-1} &= \frac34 - \frac{2k+1}{2k(k+1)} + \frac{1}{(k+1)^2-1} \\ \end{align}

Your goal is to simplify $- \frac{2k+1}{2k(k+1)} + \frac{1}{(k+1)^2-1} $ and show that it is equal to $-\frac{2(k+1)+1}{2(k+1)(k+2)}$

\begin{align} - \frac{2k+1}{2k(k+1)} + \frac{1}{(k+1)^2-1} &= - \frac{2k+1}{2k(k+1)} + \frac{1}{k(k+2)}\\ &= \frac{-(2k+1)(k+2) + 2(k+1)}{2k(k+1)(k+2)} \end{align}

Task that I left for you, simplify the numerator and you should see the solution.