Using mathematical induction to show that for any $n\ge 2$ then $\prod_{i=2}^n\bigl(1-\frac{1}{i^2}\bigr)=\frac{n+1}{2 n}$

Question

I'm trying to work through some practice problems but I've been stuck on this for god knows how long now and I've no idea where to even start. Just wondering if it would be possible for someone to break this down and go over it with me step by step.

Using mathematical induction to show that for any $n \ge 2$ then $$\prod_{i=2}^n\left(1-\frac{1}{i^2}\right)=\frac{n+1}{2 n}\tag{1}$$

Answer 1

Hint: $$\frac{n+1}{2n}\Big/\frac n{2(n-1)}=\frac{n+1}{2n}\cdot\frac{2(n-1)}n=\frac{(n+1)(n-1)}{n^2}=\frac{n^2-1}{n^2}=1-\frac1{n^2}$$

Answer 2

Base case p(2):

LHS: $$\prod_{i=2}^2 \left(1-\frac{1}{i^2}\right) = 1-\frac{1}{2^2} = \frac{3}{4}$$

RHS: $$\frac{2+1}{2\cdot2} = \frac{3}{4}$$

Now assume p(k):

$$\prod_{i=2}^k\left(1-\frac{1}{i^2}\right)=\frac{k+1}{2k}$$

$$\Rightarrow \prod_{i=2}^{k+1}\left(1-\frac{1}{i^2}\right) = \frac{k+1}{2k} \left(1-\frac{1}{(k+1)^2}\right)$$

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

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

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

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

$$=\frac{2k^3+4k^2+2k-2k}{4k^3+4k^2}$$

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

$$=\frac{k+2}{2(k+1)}$$

$$\therefore p(k) \Rightarrow p(k+1) \mbox{, so} \prod_{i=2}^n\left(1-\frac{1}{i^2}\right)=\frac{n+1}{2n} \forall n\geq 2$$

Answer 3

Check it is true for $n=2$. If it is true for $n$, then \begin{align} \prod_{i=2}^{n+1}\left(1-\frac{1}{i^2}\right) &= \prod_{i=2}^n\left(1-\frac{1}{i^2}\right) \times \left(1-\frac{1}{(n+1)^2}\right) \\&= \frac{n+1}{2n} \times \frac{n(n+2)}{(n+1)^2} \\&= \frac{n+2}{2(n+1)} \end{align}

Answer 4

Direct way.

$$\prod_{i=2}^{n}\frac{i^2-1}{i^2}=\prod_{i=2}^{n}\frac{i-1}{i} \prod_{i=2}^{n}\frac{i+1}{i}=\prod_{i=1}^{n-1}\frac{i}{i+1} \prod_{i=2}^{n}\frac{i+1}{i}=\frac{1}{2}\cdot \frac{n+1}{n}.$$