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

Question

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.

Answer 1

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 $n > 1$

Answer 2

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