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?

Without Induction:
$$
\begin{align}
\prod_{k=2}^n\left(1-\frac1{k^2}\right)
&=\prod_{k=2}^n\left[\left(1-\frac1k\right)\left(1+\frac1k\right)\right]\\
&=\prod_{k=2}^n\left(\frac{k-1}{k}\right)\prod_{k=2}^n\left(\frac{k+1}{k}\right)\\
&=\prod_{k=1}^{n-1}\left(\frac{k}{k+1}\right)\prod_{k=2}^n\left(\frac{k+1}{k}\right)\\
&=\frac12\color{#C00000}{\prod_{k=2}^{n-1}\left(\frac{k}{k+1}\right)}\cdot\frac{n+1}{n}\color{#C00000}{\prod_{k=2}^{n-1}\left(\frac{k+1}{k}\right)}\\
&=\frac{n+1}{2n}
\end{align}
$$

Comment on the Induction:
The formula for the product of $n-1$ terms is is $\frac{n+1}{2n}$.
For $n=1$, the product of $0$ terms is $1$ and that is what the formula gives.
Term $n$ is $1-\frac1{(n+1)^2}$ so the product of $n$ terms is
$$
\begin{align}
\frac{n+1}{2n}\left(1-\frac1{(n+1)^2}\right)
&=\frac{n+1}{2n}\cdot\frac{n(n+2)}{(n+1)^2}\\
&=\frac{n+2}{2(n+1)}
\end{align}
$$
which is the formula for $n$ terms.
