Mathematical Induction

**RB06** · March 1st 2009, 08:41 AM

Hi, I started this problem, got stuck, and need a hint to continue.

Prove that $1 + \frac{1}{4} + \frac{1}{9} + ... + \frac{1}{n^2} \leq 2 - \frac{1}{n}$

Base case:
$\frac{1}{1^2} = 2 - 1$
$1 =1$ so the inequality holds for $n =1$ .

Induction Hypothesis:
Assume that
$1 + \frac{1}{4} + \frac{1}{9} + ... + \frac{1}{k^2} \leq 2 - \frac{1}{k}$ , where $k$ is an arbitrary integer $\geq 1$ .

Prove that $1 + \frac{1}{4} + \frac{1}{9} + ... + \frac{1}{(k+1)^2} \leq 2 - \frac{1}{k+1} $

Here is what I got:
$1 + \frac{1}{4} + \frac{1}{9} + ... + \frac{1}{(k+1)^2} \leq 2 - \frac{1}{k} + \frac{1}{(k+1)^2}$

Any suggestion on how to continue?
Thanks in advanced.

**stapel** · March 1st 2009, 11:57 AM

. . . . . $\frac{1}{k}\, -\, \frac{1}{(k\, +\, 1)^2}\, =\, \frac{(k\, +\, 1)^2\, -\, k}{k(k\, +\, 1)^2}$

. . . . . $=\, \frac{k^2\, +\, k\, +\, 1}{k(k\, +\, 1)^2}$

. . . . . $>\, \frac{k^2\, +\, k}{k(k\, +\, 1)^2}\, =\, \frac{k(k\, +\, 1)}{k(k\, +\, 1)^2}$

. . . . . $=\, \frac{1}{k\, +\, 1}$

Use the above and the fact that, if you subtract something smaller from 2, you'll end up with something larger, to prove the final inequality.

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