How do you prove this induction?

Hello once again Jake   

Prove by induction     (n is  a nattural number)

\(1+\frac{1}{4}+\frac{1}{9}..........+\frac{1}{n^2}<2-\frac{1}{n}\\\)

Step 1

Prove true for n=1

LHS = 1/1^2 = 1

RHS = 2-1/1 = 1 LHS

Therefore true for n=1

Step 2

Assume true for n=k

\(1+\frac{1}{4}+\frac{1}{9}..........+\frac{1}{k^2}<2-\frac{1}{k}\\\)

Now prove true for n=k+1

that is, prove

\(1+\frac{1}{4}+\frac{1}{9}..........+\frac{1}{k^2}+\frac{1}{(k+1)^2}<2-\frac{1}{k+1}\\ 1+\frac{1}{4}+\frac{1}{9}..........+\frac{1}{k^2}+\frac{1}{(k+1)^2}<2-\frac{k(k+1)}{k(k+1)^2}\\ 1+\frac{1}{4}+\frac{1}{9}..........+\frac{1}{k^2}+\frac{1}{(k+1)^2}<2-\frac{k^2+1}{k(k+1)^2}\\ \\ LHS<(2-\frac{1}{k})+\frac{1}{(k+1)^2}\\ LHS<2-\left(\frac{1}{k}-\frac{1}{(k+1)^2}\right)\\ LHS<2-\left(\frac{(k+1)^2}{k(k+1)^2}-\frac{k}{k(k+1)^2}\right)\\ LHS<2-\left(\frac{k^2+k+1}{k(k+1)^2}\right)\\ LHS<2-\frac{k^2+1}{k(k+1)^2}-\frac{k}{k(k+1)^2}\\ therefore\\ LHS<2-\frac{k^2+1}{k(k+1)^2}\\ LHS<2-\frac{1}{k+1}\\ \)

Therefore, if true for n=k then it must be true for n=k+1

Step 3

Since true for n=1, it must be true for n=2

Since true for n=2 it must be true for n=3, n=4,n=5, etc.  

Therefore true for al n   where   \(n\in N\)

anyone?

wow that must've took a while thanks so much! i needed this for my homework my teacher's not a forgiving person.

Yes well, I hope you learn from it!

I want to help you learn, it was not my intention just to do your homework for you!!