Proof by induction $\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} < 2$ 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.

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