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Mathematical Induction Example Proof statement

Mathematical Induction Example Proof statement

Problem Statement:

Using Mathematical Induction, prove that  for n > 1, the following formula holds true:

 

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

Mathematical Proof:

First of all, we will prove for the base case n=2.

LHS  =  \LARGE \left ( 1 - \frac{1}{2.2} \right ) = \frac{3}{4}

 

RHS = \LARGE \frac{ 2 + 1}{2.2} = \frac{3}{4}

 

Both LHS and RHS ( Left hand side and Right hand side ) are equal. So its is proved for the base case.

Let us assume that for P(k)  for k ≥ 2 be a arbitrarily chosen integer such that the equation holds true.
\LARGE \prod_{i=2}^{k}\left ( 1 - \frac{1}{i^{2}} \right ) = \frac{k+1}{2k}
Using that we prove that for P(k+1) , the equation holds true, such that :
\LARGE \prod_{i=2}^{k+1}\left ( 1 - \frac{1}{i^{2}} \right ) = \frac{(k+1) +1}{2(k+1)}
Let us multiply the k+1 term using the P(k) , we get:
\LARGE \prod_{i=2}^{k}\left ( 1 - \frac{1}{i^{2}} \right ).\left ( 1-\frac{1}{(k+1)^{2}} \right ) = \frac{k+1}{2k}.\left ( 1-\frac{1}{(k+1)^{2}} \right )
Simplify and get to the form of P(k+1)
\LARGE \prod_{i=2}^{k+1}\left ( 1 - \frac{1}{i^{2}} \right ) = \frac{k+1}{2k}.\left ( 1-\frac{1}{(k+1)^{2}} \right )
Expand the RHS and get to the form of p(k+1), we get
RHS =  \LARGE \frac{k+1}{2k} -\frac{k+1}{2k(k+1)^{2}}
=     \LARGE \frac{k+1}{2k} - \frac{1}{2k(k+1)}
=    \LARGE \frac{(k+1)^2 -1}{2k(k+1)}
=    \LARGE \frac{k^{2}+2k}{2k(k+1)}
Cancel the k, we get
=  \LARGE \frac{k+2}{2(k+1)}
= \LARGE \frac{(k+1)+1}{2(k+1)}
We proved for the P(k+1) using the P(k)
\LARGE \prod_{i=2}^{k+1}\left ( 1 - \frac{1}{i^{2}} \right ) = \frac{(k+1)+1}{2(k+1)}
Hence proved that:
\LARGE P(k) \Rightarrow P(k+1)

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