How to Complete the Inductive Step in a Series Inequality Proof?

[sbc824]

Homework Statement

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

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)

 

[SammyS]

Homework Statement

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

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)

To clarify matters:

I take it that you need to prove (by induction) that:

\displaystyle <br /> \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.

 

[sbc824]

To clarify matters:

I take it that you need to prove (by induction) that:

\displaystyle <br /> \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.

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.

 

[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.

 

[SammyS]

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.

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.