The method I used: Mathematical induction
I am aware that $\sum_{k=1}^{n} \frac{1}{k^2} = \frac{\pi^2}{6}$
Wanted to find a general proof than going from that level of specificity.
It is trivial to prove the case for $n=1$
We need to prove $$\sum_{k=1}^{p} \frac{1}{k^2} < 2 \implies \sum_{k=1}^{p+1} \frac{1}{k^2} < 2 \tag{1}$$
Assuming $$\sum_{k=1}^{p} \frac{1}{k^2} < 2 \tag{2}$$ Then $$\sum_{k=1}^{p+1} \frac{1}{k^2} < 2 + \frac{1}{(p+1)^2} \tag{3}$$
We know, $$p \geq 1 $$ $$ p+1 \geq 2 $$ $$ (p+1)^2 \geq 4 $$ $$ \frac{1}{(p+1)^2} \leq \frac{1}{4} $$ $$ 2+\frac{1}{(p+1)^2} \leq 2+\frac{1}{4} \tag{4}$$
By inequalities 3 and 4: $$\sum_{k=1}^{p+1} \frac{1}{k^2} < 2 + \frac{1}{(p+1)^2} \leq 2+\frac{1}{4} \tag{5}$$
Therefore we get $$\sum_{k=1}^{p} \frac{1}{k^2} < 2 \implies \sum_{k=1}^{p+1} \frac{1}{k^2} < 2 +\frac{1}{4} \tag{6}$$ Anything I've missed to reach statement 1?
