Induction to Prove Series Inequality

[tajmorton]

Homework Statement

Show $$\sum_{i=1}^n \frac{1}{i^2}$$ $$\leq$$ $$2 - \frac{1}{n}$$ 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 $$\leq$$ 2 - 1/1
(works)

Inductive Hypothesis:
$$\sum_{i=1}^n \frac{1}{i^2} = S_{j}$$
$$S_{j} \leq 2 - \frac{1}{j}$$ for all j = 0, 1, ... n

Inductive Step:
$$S_{n+1} = S_{n} + \frac{1}{{(n+1)}^2}$$

Using inductive assumption:
$$S_{n} + \frac{1}{{n+1}^2} \leq 2 - \frac{1}{n} + \frac{1}{{(n+1)}^2}$$

My plan is to show that
$$2 - \frac{1}{n} + \frac{1}{{(n+1)}^2} \leq 2 - \frac{1}{n+1}$$,
since by the inductive hypothesis we have
$$S_{n} \leq 2 - \frac{1}{n}$$.

So, if I can show that the first inequality is true, then I think the following should be true:
$$S_{n} \leq 2 - \frac{1}{n} + \frac{1}{{(n+1)}^2} \leq 2 - \frac{1}{n+1}$$

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

 

[CompuChip]

Sounds good, and perhaps this would help;

$$\frac{1}{n} - \frac{1}{(n+1)^2} = \frac{1}{n+1} \left( \frac{n+1}{n} - \frac{1}{n + 1} \right) = \frac{1}{n+1} \left( 1 + \frac{1}{n} - \frac{1}{n + 1} \right)$$