How to show the following limit Let $f:\mathbb{N^*} \to \mathbb{N^*}$ an injective function. Show that $\displaystyle \sum _{n=0}^{+\infty}\frac{f(n)}{n^2}=+\infty$. I tried to :

*

*Show that $f(n) \ge n$ for all $n$, but i think i've found that's false. ($f(1)=2$ $f(2)=1$ for example)


*Let $(S_n)_n$ a sequence such that $S_n= \frac{f(n)}{n^2}$, then using some criterion like Alembert criterion's, yet it didn't work.
Any help, please !
 A: For all $n \geq 1$, let $$S_n = \sum_{k=1}^n \frac{f(k)}{k^2}$$
You have that
$$S_{2n}-S_n = \sum_{k=n+1}^{2n} \frac{f(k)}{k^2} \geq \frac{1}{(2n)^2} \sum_{k=n+1}^{2n} f(k) \geq \frac{1}{4n^2}(1+2+...+n)=\frac{n(n+1)}{8n^2}\geq \frac{1}{8}$$
(justification of $\sum_{k=n+1}^{2n} f(k) \geq 1+2+...+n$ : the sum is made by $n$ distinct integers - because $f$ is injective -, so it is greater or equal to the smallest sum of $n$ distinct integers you can have, which is $1+2+...+n$.)
So $S_{2n}-S_n$ does not tend to $0$, so $(S_n)$ does not converge.
A: Introduce $S_n=\sum_{k=1}^n \frac{f(k)}{k^2}$. Notice that the first $n$ values of $f(n)$ must be different because of injection. The minimum value of $S_n$ is obtained when the first $n$ values of $f(n)$ are $1,2,3\dots n$.
First show that for any $k,l$ such that $k<l\le n$:
$$\frac{k}{k^2}+\frac{l}{l^2}<\frac{l}{k^2}+\frac{k}{l^2}$$
This is obvious, because you can easily show that it is equivalent to:
$$kl^2+lk^2<l^3+k^3$$
$$0<l^2(l-k)-k^2(l-k)$$
$$0<(l^2-k^2)(l-k)$$
...which is true for $l>k$. It means that the minimum value of $S_n$ is achieved when $f(k)=k$ for $k=1,2,\dots,n$
It means that:
$$S_n\ge S_{n_{min}}=\sum_{k=1}^n \frac{k}{k^2}=\sum_{k=1}^n \frac{1}{k}=H_n$$
...where $H_n$ is the $n$-th harmonic number.
It means that:
$$\lim_{n\to \infty}S_n\ge\lim_{n\to \infty}H_n=\infty$$
$$\sum_{n=1}^{\infty}\frac{f(n)}{n^2}=\infty$$
