For $n \geq 2$, prove that $(1- \frac{1}{4})(1- \frac{1}{9})(1- \frac{1}{16})...(1- \frac{1}{n^2}) = \frac{n+1}{2n}$ For $n \geq 2$ prove that $(1- \frac{1}{4})(1- \frac{1}{9})(1- \frac{1}{16})...(1- \frac{1}{n^2}) = \frac{n+1}{2n}$
We need to use induction. 
The Principle of Mathematical Induction, Theorem 4.2.1, states that $n_0 \in \mathbb{Z}$. For each integer $n \geq n_o$, let $P(n)$ be a statement about $n$. Suppose that the following two statements are true:


*

*$P(n_o)$

*$( \forall n \geq n_o)[P(n) \rightarrow P(n+1)]$


Then, for all integers $n \geq n_o$, that statement $P(n)$ is true.
Suppose, 
$(1- \frac{1}{4})(1- \frac{1}{9})(1- \frac{1}{16})...(1- \frac{1}{n^2}) = \frac{n+1}{2n}$, the equation, is true for some natural number $k$.
$\frac{1}{2^2})*...*(1- \frac{1}{k^2}) = \frac{k+1}{2k}$
The induction hypothesis $P(k) \rightarrow P(k+1)$ must hold for any natural number $k, P(k),$ from the basis and inductive step.
$(\frac{1}{2^2})*...*(1- \frac{1}{k^2})*(1- \frac{1}{(k+1)^2}) = \frac{k+1}{2k}(1- \frac{1}{(k+1)^2})$
$\leftrightarrow$ $\frac{k+1}{2k}(\frac{(k+1)^2}{(k+1)^2}- \frac{1}{(k+1)^2})$
$\leftrightarrow$ $(\frac{(k+1)}{2k})( \frac{(k+1)^2-1}{(k+1)^2})$
$\leftrightarrow$ $( \frac{(k^2+2k+1-1}{(2k)(k+1)})$
$\leftrightarrow$ $( \frac{(k^2+2k}{(2k)(k+1)})$
$\leftrightarrow$ $( \frac{k+2}{(2)(k+1)})$
We need to demonstrate that 
$(\frac{1}{2^2})*...*(1- \frac{1}{n^2}) = \frac{n+1}{2n}$
holds true for $n \geq 2$
The question is do I just substitute a number that is greater than two after I did the induction?

Here is an answer that uses logarithmic functions.
Let
$$
P_n=\prod_{k=2}^{n}\left(1-\frac{1}{k^2}\right)
$$
and
$$
S_n=\log P_n.
$$
Then
\begin{eqnarray}
S_n&=&\log\prod_{k=2}^{n}\left(1-\frac{1}{k^2}\right)\\
&=&\sum_{k=2}^{n}\log\left(1-\frac{1}{k^2}\right)\\
&=&\sum_{k=2}^{n}\log\left(\frac{k^2-1}{k^2}\right)\\
&=&\sum_{k=2}^{n}\log\left[\left(\frac{k-1}{k}\right)\left(\frac{k+1}{k}\right)\right]\\
&=&\sum_{k=2}^{n}\left[\log\left(\frac{k-1}{k}\right)+\log\left(\frac{k+1}{k}\right)\right]\\
&=&\sum_{k=2}^{n}\left[\log(k-1)-\log(k)\right]+\sum_{k=2}^n\left[\log(k+1)-\log(k)\right]\\
&=&\sum_{k=1}^{n-1}\log(k)-\sum_{k=2}^{n}\log(k)+\sum_{k=3}^{n+1}\log(k)-\sum_{k=2}^{n}\log(k)\\
&=&\log(1)+\sum_{k=2}^{n-1}[\log(k)-\log(k)]-\log(n)+\log(n+1)+\sum_{k=3}^{n}[\log(k)-\log(k)]-\log(2)\\
&=&0+0-\log(n)+\log(n+1)+0-\log(2)\\
&=&\log\left(\frac{n+1}{2n}\right).
\end{eqnarray}
Since
$$
S_n=\log P_n=\log\left(\frac{n+1}{2n}\right),
$$
it follows that
$$
P_n=\prod_{k=2}^{n}\left(1-\frac{1}{k^2}\right) =\frac{n+1}{2n}
$$
