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Proof by induction $\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} < 2 \ \ \ n \in \mathbb{N}$

So for $n=1$

$$ 1 < 2$$

For $n > 1$

Assumption: $$\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} < 2$$

Hypothesis (inductive step):

$$\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} + \frac{1}{(n+1)^2} < 2$$

So using assumption and hypothesis I have:

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

So then: $$ \frac{1}{(n+1)^2} > 0 $$ which is always true

I was told it's relatively "hard" one. Thus I think I made sth stupid here.

tomtom
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    You did not prove the induction step. "Assumption" and "Hypothesis" basically mean the same thing. So you assumed the $n$'th step, and then you assumed again the $(n+1)$'th step, when in fact you needed to prove it. That's not a valid proof. – uniquesolution Oct 21 '15 at 21:21
  • There's some room to play with, here. You may be intrested to see that the smallest value this could be proved for is $\pi^2/6$ instead of $2$. It will be difficult to get anywhere near that with induction, though. https://en.wikipedia.org/wiki/Basel_problem – Niklas Oct 21 '15 at 21:47
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    @Niklas On can get somewhat near with elementary means. – Daniel Fischer Oct 21 '15 at 21:58
  • @DanielFischer Thanks. And surprisingly (to me) these elementary means actually include induction. – Niklas Oct 21 '15 at 22:05

2 Answers2

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HINT: It is easier to prove by induction this: $\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} < 2 - \frac{1}{n}$ for $n > 1$

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inequility： $\frac{2}{n^2}\leq \frac{2}{n^2-1}=\frac{1}{n-1}-\frac{1}{n+1}$

chenyuandong
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