Proving $\sum_{i=1}^n (1-\frac{1}{(i+1)^2}) = \frac{n+2}{2n+2}$ using induction. My textbook has the following question:

Prove the follwing statement using induction for all natural numbers $n$
$$(1- \frac{1}{4})+(1- \frac{1}{9})+.......+(1- \frac{1}{(n+1)^2})=\frac{n+2}{2n+2}$$

So, I check both the sides for $n=1$. In that case LHS=RHS=$\frac{3}{4}$.
Now  I assume the statement to be true for $n=k$ which gives
$$
\sum_{i=1}^k (1-\frac{1}{(i+1)^2}) = \frac{k+2}{2k+2}
$$
Now I evaluate the original statement for $n=k+1$ which leaves us with the
$$ LHS= \frac{3k^3+16k^2+26k+14}{2(k+1)(k+2)^2} $$
And we have to prove this LHS to be equal to RHS which is
$$
RHS= \frac{k+3}{2k+4}
$$
But these (new) LHS and RHS don't seem to be equal. And hence I am not able to complete the proof.
How should I proceed?
A solution without induction is also welcome.

Book: Comprehensive Algebra VOL-1
Author: Vinay Kumar
Publisher: McGraw Hill Education.
 A: The formula you are trying to prove is wrong. The actual correct formula is
$$ \prod_{i=1}^n \left(1-\frac{1}{(i+1)^2}\right)=\frac{n+2}{2n+2}=\frac{n+2}{2(n+1)} $$
For $n=1$, this is easy to see.
$$\left(1-\frac{1}{(1+1)^2}\right)=\frac{3}{4}=\frac{1+2}{2\cdot 1+2}$$
For the induction step:
$$\prod_{i=1}^{n+1} \left(1-\frac{1}{(i+1)^2}\right)=\left(1-\frac{1}{(n+2)^2}\right)\prod_{i=1}^{n} \left(1-\frac{1}{(i+1)^2}\right)=\left(1-\frac{1}{(n+2)^2}\right)\cdot \frac{n+2}{2n+2}$$
Further computing leads to
$$\left(1-\frac{1}{(n+2)^2}\right)\cdot \frac{n+2}{2n+2}=\frac{n+2}{2(n+1)}-\frac{1}{2(n+1)(n+2)}=\frac{n^2+4n+4-1}{2(n+1)(n+2)}=\frac{n^2+4n+3}{2(n+1)(n+1+1)}$$
Finally by dividing by $n+1$ follows
$$\frac{n^2+4n+3}{2(n+1)(n+1+1)}=\frac{n+1 +2}{2(n+1+1)}$$
which is what we wanted to show.
A: So the claim is
$$\sum_{i=1}^{n} (1-\frac{1}{(i+1)^2}) = \frac{n+2}{2n+2}$$
Base case is true, now assuming that it is true for all naturals less than $n+1$ we will show that it is true for $n+1$.
$$\sum_{i=1}^{n+1} (1-\frac{1}{(i+1)^2}) = \frac{n+3}{2n+4}$$
is to be proved
Instead of working with summation,we will estimate difference.
Obviosuly
$$\frac{n+3}{2n+4}- \frac{n+2}{2n+2} = 1-\frac{1}{(n+2)^2}$$
if the formula is correct (and also, using inductive hypothesis)
But
$$\frac{n+3}{2n+4}- \frac{n+2}{2n+2} = \frac{1}{2} \left( \frac{(n+3)(n+1)-(n+2)^2}{(n+2)(n+1)} \right) = $$
$$ = -\frac{1}{2(n+1)(n+2)}$$
which is obviously wrong
A: For $n=2$ we have
$$\sum_{i=1}^{2}(1-\frac{1}{(i+1)^{2}})=(1-\frac{1}{4})+(1-\frac{1}{9})=\frac{59}{36}$$ but $$\frac{n+2}{2n+2}\big|_{n=2}=\frac{2+2}{2(2)+2}=\frac{2}{3}\neq\frac{59}{36}.$$
