# Inductive proof help on last step

1. Sep 2, 2012

### sbc824

1. The problem statement, all variables and given/known data

i=1 Sigma n (1/i2) <= 2 - (1/n)

3. The attempt at a solution

I've done the basic step and assumption step...little stuck on the inductive step

So far I have...

show 1 + 1/4 + 1/9 + 1/16 +...+ (1/k2) + (1/(k+1)2) <= 2 - (1/k+1)

2. Sep 3, 2012

### SammyS

Staff Emeritus
To clarify matters:

I take it that you need to prove (by induction) that:
$\displaystyle \sum_{i=1}^{n}\frac{1}{i^2}\le2-\frac{1}{n}\ .$​

Is that correct?

So, you have assumed that 1 + 1/4 + 1/9 + 1/16 +...+ (1/k2) ≤ 2 - (1/k) ,

and you need to show that 1 + 1/4 + 1/9 + 1/16 +...+ (1/k2) + (1/(k+1)2) ≤ 2 - (1/(k+1)) .

Is that correct?

What have you tried, in this effort?

BTW: Please learn to use parentheses, so that your mathematical expressions say what you mean for them to say.

3. Sep 4, 2012

### sbc824

Yes, that is correct...I've done equality inductive proofs, but have not encountered less than or greater than type proofs...so I'm not sure how to begin.

4. Sep 4, 2012

### dot.hack

The idea is almost exactly the same as equality. Starting from where you left off you have $\sum_{i=1}^{n}\frac{1}{i^2} + \frac{1}{n+1}$≤ 2- $\frac{1}{n+1}$
See what happens if you combine like terms and use what you already know about $\sum_{i=1}^{n}\frac{1}{i^2}$ to help you out.

5. Sep 4, 2012

### SammyS

Staff Emeritus
Let's see:

You are assuming that some k,
1 + 1/4 + 1/9 + 1/16 +...+ 1/k2 ≤ 2 - (1/k) .​

Adding 1/(k+1)2 to both sides gives you that
1 + 1/4 + 1/9 + 1/16 +...+ 1/k2 + 1/(k+1)2 ≤ 2 - (1/k) +1/(k+1)2 .​

So, if you can show that 2 - (1/k) +1/(k+1)2 ≤ 2 - 1/(k+1), then you should be able to complete the proof.

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