Question: Prove that there is no bijective function $g:\Bbb N\to \Bbb N$ , such that $ $$\sum_{n=1}^\infty \frac{g(n)}{n^2}$$ < \infty$ .(This question is fromTIFR Mathematics, 2021)
My attempt:
Since $g:\Bbb N\to \Bbb N$ is bijective, so $g(n_{1}) \neq g(n_2)$ $\Rightarrow n_1 \neq n_2$.
Now g(n) gives a natural number . So we can write $g(n)= n^{x_n}$, $\forall$ n>1 ; where $ x_n\in \Bbb R^+ $
Therefore, $ \frac{g(n)}{n^2}$$=$$\frac{1}{n^{(2-{x_n})}} > \frac {1}{n^2}$ .
Now , $\sum_{n}\frac {g(n)}{n^2}$ =$\sum_{n=2}\frac {1}{n^{(2-x_n)}}+g(1)$ > $\sum_{n}\frac{1}{n^{(2-max.{{x_n}})}}+g(1)-1$ .
Now ,$\Biggr(\sum_{n}\frac{1}{n^{(2-max.{x_n})}}+g(1)-1\Biggr)$ is divergent since $(2-max.\{x_n\}) \le 1$ because there exist a pair $(m,n)\in\Bbb N ×\Bbb N$ such that $x_i=\log_n m \ge 1$ Therefore $max.\{x_n\}\ge 1$
and so is $\sum_{n}\frac {g(n)}{n^2}$ is divergent.
Hence $\sum_{n}\frac {g(n)}{n^2} \not\lt \infty$ .
Now I'm not sure whether $\sum_{n}\frac {g(n)}{n^2} \not\lt \infty$ can be written like this or not when the series is divergent . For example we know $\sum_{n}\frac{1}{n}<\sum_{n}n$ , $\forall n\in\Bbb N $ here both the series of L.H.S and R.H.S are divergent but if we follow the inequality ,$\sum\frac{1}{n}<\infty$ .
Is my attempt to this problem correct ? If not please help to prove that.