# HELP WITH PROOFS!!! Prove that 1 + 1/4 + 1/9 +....1/n^2 ≤ 2- (1/n) for every positive integer n.?

Prove that 1 + 1/4 + 1/9 +....1/n^2 ≤ 2- (1/n) for every positive integer n.

Please go step by step because i dont understand where to go or even how to do this problem and state what kind of proof you are using (ie. direct, counterexample, contrapositive, basic induction, etc.)

### 3 Answers

- NancyLv 67 years ago
using induction:

if n = 1, 1/1^2 = 1 <= 2 - 1/1 = 1 so this is true for n = 1

assume that the statement is true for n = k, 1 + 1/4 + 1/9 + ... + 1/k^2 <= 2 - (1/k)

add 1/(k + 1)^2 to both sides: 1 + 1/4 + 1/9 + ...+1/k^2 + 1/(k + 1)^2 <= 2 - (1/k) + 1/(k + 1)^2 <=

2 - [(k + 1)^2 - k]/k(k + 1)^2 <= 2 - [k^2 + k + 1]/k(k + 1)^2 = 2 - (k(k + 1)/k(k + 1)^2 - 1/k(k + 1) =

2 - 1/(k + 1) - 1/(k(k + 1)^2 < 2 - 1/(k + 1)

- kbLv 77 years ago
Here is a non-inductive direct proof.

1 + 1/2^2 + 1/3^2 + ... + 1/n^2

≤ 1 + 1/(1*2) + 1/(2*3) + ... + 1/((n-1)n)

= 1 + (1 - 1/2) + (1/2 - 1/3) + ... + (1/(n-1) - 1/n), by partial fractions

= 1 + (1 - 1/n), since all other terms cancel in pairs

= 2 - 1/n.

I hope this helps!

- degeorgeLv 43 years ago
i'd be starting up from the inductive step. ? k2^ok + (ok+a million)2^(ok+a million) : (ok-a million)2^(ok+a million) + 2 + (ok+a million)2^(ok+a million) : ok 2^(ok+a million) - 2^(ok+a million) + 2 + ok 2^(ok+a million) + 2^(ok+a million) : 2k 2^(ok+a million) + 2 : ok 2^(ok+a million + a million) + 2 : ok 2^(ok+2) + 2 desire this allows!!!!!