Induction problem

**tttcomrader** · November 18th 2006, 09:05 AM

Q: Prove the set of { $S = n \epsilon <b>N</b> : \frac{1}{1^2} + \frac{1}{2^2} + ... + \frac{1}{n^2} <= 2 - \frac{1}{n}$ } is inductive.

P.S. Are there any website or program in which I can learn more about the math coding in here? I had been spending hours inputing simply math operations. For example, I had just typed out my progress of work so far in this problem, took me about half an hour, but after I hit save, it just disappeared, did not update it, not anything. So I'm not going to type out my progress, sorry!

Thank you.

**earboth** · November 18th 2006, 10:51 AM

Originally Posted by tttcomrader

Q: Prove the set of { $S = n \epsilon <b>N</b> : \frac{1}{1^2} + \frac{1}{2^2} + ... + \frac{1}{n^2} <= 2 - \frac{1}{n}$ } is inductive....

Hello,

If I understand your problem right, you want to prove this property by mathematical induction(?):

I assume that you know how to do MI:

1. Prove for n = 1: $\frac{1}{1^2}=2-\frac{1}{1}$ is true.

2. Assume that $\frac{1}{1^2} + \frac{1}{2^2} + ... + \frac{1}{n^2} <= 2 - \frac{1}{n}$ is true.

3. Prove that the step from n to (n+1) is true under the assumption:

$\frac{1}{1^2} + \frac{1}{2^2} + ... + \frac{1}{n^2}+ \frac{1}{(n+1)^2} <= 2 - \frac{1}{n}+ \frac{1}{(n+1)^2}$

$<= 2 - \frac{(n+1)^2-n}{n \cdot (n+1)^2}$ . Now we add $\frac{1}{n \cdot (n+1)^2}$ which is possible because of the <= sign!

$<= 2 - \frac{(n+1)^2-n-1}{n \cdot (n+1)^2}$

$<= 2 - \frac{(n+1)^2-(n+1)}{n \cdot (n+1)^2}$

$<= 2 - \frac{(n+1)((n+1)-1)}{n \cdot (n+1)^2}$

$<= 2 - \frac{(n+1)n}{n \cdot (n+1)^2}$ . Cancel out the equal factors:

$<= 2 - \frac{1}{n+1}$

Thus this property is true for all $n \in \mathbb{N}$

EB

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