- #1

tajmorton

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## Homework Statement

Show [tex]\sum_{i=1}^n \frac{1}{i^2}[/tex] [tex]\leq[/tex] [tex]2 - \frac{1}{n}[/tex] with induction on n.

I'm pretty rusty on induction (not that I was very good at it to being with), so I mostly wanted to know if I'm on the right track, and if this is a way towards a valid proof.

## Homework Equations

## The Attempt at a Solution

Base Step (n=1): 1 [tex]\leq[/tex] 2 - 1/1

(works)

Inductive Hypothesis:

[tex]\sum_{i=1}^n \frac{1}{i^2} = S_{j}[/tex]

[tex]S_{j} \leq 2 - \frac{1}{j}[/tex] for all j = 0, 1, ... n

Inductive Step:

[tex]S_{n+1} = S_{n} + \frac{1}{{(n+1)}^2}[/tex]

Using inductive assumption:

[tex]S_{n} + \frac{1}{{n+1}^2} \leq 2 - \frac{1}{n} + \frac{1}{{(n+1)}^2}[/tex]

My plan is to show that

[tex]2 - \frac{1}{n} + \frac{1}{{(n+1)}^2} \leq 2 - \frac{1}{n+1}[/tex],

since by the inductive hypothesis we have

[tex]S_{n} \leq 2 - \frac{1}{n}[/tex].

So, if I can show that the first inequality is true, then I think the following should be true:

[tex]S_{n} \leq 2 - \frac{1}{n} + \frac{1}{{(n+1)}^2} \leq 2 - \frac{1}{n+1}[/tex]

Would that be a valid proof?

Apologies about the poor TeX, I tried to fix some of the problems I saw, but they never updated in my browser (cache?).

Any pointers would be appreciated... Thanks!

- Taj