Proof by induction help please!? ((1-(1/2^2)) * (1-(1/3^2)) * 1-(1/4^2)) ... (1-(1/n^2)) = (n+1)/2nI'm specifically having trouble after the base case. Any help would be appreciated! Thanks!

1 Answer

rcmath's profile pic

rcmath | High School Teacher | (Level 1) Associate Educator

Posted on

From the given, we notice that the smallest value n can take is 2. 

So when n=2, we have `1-1/(2^2)=1-1/4=3/4`    and `(2+1)/[2*2]=3/4`

Hence the statement is true for n=2. Now suppose that the statement is true for n, we need to prove it for n+1. In other words we need to prove the following equality:


Let's simplify the right hand side = `(n+2)/[2(n+1)]`

Since we are assuming that the statement is true for n, then we know that `(1-1/[2^2])*(1-1/[3^2])*...*(1-1/[n^2])=(n+1)/[2n]`

If we substitue the following in the Left Hand side of the equation we are proving we get






`[(n+1)*n*(n+2)]/[2n(n+1)^2]=` If we simplify, we get

`(n+2)/[2(n+1)]` Which is equal to the right hand side.

Hence the statement is proved by induction.