Find the number of bijective function $g(n):\mathbb{N}\to \mathbb{N}$ 
Find the number of bijective function  $g(n):\mathbb{N}\to \mathbb{N}$ such that it's satisfies $\sum_{n=1}^{\infty} \frac{g(n)}{n^2}<\infty$


I think  there is no such bijective function exists , suppose $g(n)=n$ then $\frac{g(n)}{n^2}=\frac{n}{n^2}=\frac{1}{n}$ , but $\sum_{n=1}^{\infty}\frac{1}{n} $ is  diverges .
Now again $\sum_{n=1}^{\infty}\frac{1}{n} $ is converges if we omit the term $n$ whose last entries is $9$. But then don't understand how to construct such bijective function .(source:: https://math.stackexchange.com/a/1343048/746904)

This question is came in TIFR GS-2021 .

Abel summation  can be used here.
Let $G(n)$ denote the sum $g(1) + \dots + g(n)$. We then have, for any fixed $M$, $$\sum_{n = 1}^M \frac{g(n)}{n^2} > \sum_{i = 1}^{M - 1} G(i)\left(\frac 1{i^2} - \frac 1 {(i + 1)^2}\right).$$
Using the fact that $g$ is a bijection from $\Bbb N$ to $\Bbb N$, we know that $G(n) \geq 1 + 2 + \dots + n = \frac{n(n + 1)}2$ for any $n$. Putting this into the right hand side of the above equality will give you an estimation that diverges to infinity when $M$ tends to infinity.

As can be seen from the proof, in fact we only used the injectivity of $g$.
