Does there exist a bijective function $g:\mathbb{N}\to \mathbb{N}$ such that $\sum_{n=1}^{\infty}\frac{g(n)}{n^2}<\infty$
My approach
Suppose it does then the sequence $(\frac{g(n)}{n^2})_{n\in \mathbb{N}}$ converges to zero.
Then for any $\epsilon>0\ $ we get an $ \ m\in \mathbb{N}$ such that $ \ 0<\frac{g(n)}{n^2}<\epsilon, \ \forall\ n\geq m $
$g(n)<\epsilon n^2, \forall\ n\geq m$
can't proceed from here...
more thoughts
Observations:
- For g(n)=1 constant function the series is convergent, can't find any more such which for good is bijective
- Taking the bijective function g(n)=n we get that the series becomes the harmonic series which is divergent. That also won't work
But those gives no conclusion, any idea ?