Prove for all integers n>1 that 1+1/4+1/9+...+1/n² < 2 - 1/n?

Question

Can't figure out how to solve this one:
Prove for all integers n>1 that
1+1/4+1/9+...+1/n² < 2 - 1/n

I believe that this is done using mathematical induction

Answer 1

For n=2

1+1/4 = 1.25 < 1.5 = 2 - 1/2

Let's assume for n>2 and prove for n+1

1+1/4+...+1/n^2 + 1/(n+1)^2 < 2-1/n + 1/(n + 1)^2 =
= 2 - [(n + 1)^2 - n]/[n(n + 1)^2] =
= 2 - (n^2 + 2n + 1 - n)/[n(n + 1)^2] =
= 2 - (n^2 + n + 1)/[n(n + 1)^2] < 2 - (n^2 + n)/[n(n + 1)^2 =
= 2 - (n + 1)/(n + 1)^2 = 2 - 1/(n + 1)

Answer 2

yes, induction works. 1<2, so this is the base case. If Sum_{i=1}^n 1/i^2 < 2-1/n for some n, then:

Sum_{i=1}^{n+1} 1/i^2 < 2 - 1/n + 1/(n+1)^2.

But:

2 - 1/n + 1/(n+1)^2 = 2 + (n - (n+1)^2)/(n(n+1)^2

= 2 + (-n^2 - n -1)/(n(n+1)^2) < 2 - n(n+1)/(n(n+1)^2) = 2 - 1/(n+1).

This completes the inductive step.

Answer 3

definite, induction works. a million<2, so it is the backside case. If Sum_{i=a million}^n a million/i^2 < 2-a million/n for some n, then: Sum_{i=a million}^{n+a million} a million/i^2 < 2 - a million/n + a million/(n+a million)^2. yet: 2 - a million/n + a million/(n+a million)^2 = 2 + (n - (n+a million)^2)/(n(n+a million)^2 = 2 + (-n^2 - n -a million)/(n(n+a million)^2) < 2 - n(n+a million)/(n(n+a million)^2) = 2 - a million/(n+a million). This completes the inductive step.