Show that the sum is between (1,2)

Question

Show that the sum $$1+\frac{1}{2^2}+\frac{1}{3^2}+\dots+ \frac{1}{2022^2}$$ is between $(1,2)$, I mean is bigger than $1$ and less than $2$.

What I tried: honestly I never knew or understood how to do this type of exercise, but I have some ideas:

I thought about a formula: $$1/k^2<1/(k-1)×k$$ I can use this one, but I don't think it could help, because I'm only going to prove something with this, I can't calculate. Should I use the formulas with the terms of the sum, where I find out how many numbers there are in my sum? Any suggestion will help, thanks!

Answer 1

Clearly you have: $$1<\sum_{k=1}^{2022}\frac{1}{k^2}<\sum_{k=1}^{\infty}\frac{1}{k^2} $$ Now note that: $$\sum_{k=1}^{\infty}\frac{1}{k^2}<\sum_{k=1}^{\infty}\frac{2}{k(k+1)}=2\sum_{k=1}^{\infty}\left(\frac{1}{k}-\frac{1}{k+1}\right)=2$$

Answer 2

$$1 < \sum_{n=1}^{2022} \frac{1}{n^2} < \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}<2 $$