for a differentiable function $f$ in $(0,\infty)$ and $ 0for a differentiable function $f$  in  $(0,\infty)$ and $  0<f'(x)<\frac{1}{x^{2}}  $ I need to prove that $\lim_{n\to\infty}(f((n+1)^{2})-f(n^{2}))=0$.
First thing that came to my mind is uniform continuity because the derivative is bounded, but how can it serves me here? 
Thank you.
 A: We get the result thanks to the inequalities
$$0\leq \int_{n^2}^{(n+1)^2}f'(t)dt=f((n+1)^2)-f(n^2) \leq \int_{n^2}^{(n+1)^2}\frac 1{x^2} dx = \frac{-1}x\mid_{n^2}^{(n+1)^2}=\frac 1{n^2}-\frac 1{(n+1)^2}.$$ 
A: The mean value theorem can serve you here.
A: The following steps lead to a solution:
(1) Note the Mean Value Theorem in this context:
If $f$ is a differentiable function on $(0,\infty)$, then for all $a,b\in (0,\infty)$, $a<b$, there exists $c$ such that $a<c<b$ and:
$f(b)-f(a)=f'(c)(b-a).$
(2) Deduce that for all positive integers $n$, we have $f((n+1)^2)-f(n^2)=f'(c_n)((n+1)^2-n^2)$ for some real number $c_n$ such that $n^2<c_n<(n+1)^2$. 
(3) Show that $(n+1)^2-n^2=2n+1$ and $\frac{1}{c_n}<\frac{1}{n^2}$ for all positive integers $n$.
(4) Deduce that $\left|f((n+1)^2)-f(n^2)\right|=\left|f'(c_n)\right|\left|(2n+1)\right|<\frac{2n+1}{c_n^2}<\frac{2n+1}{n^4}$.
(5) Finally, conclude that $\lim_{n\to\infty} \left[f((n+1)^2)-f(n^2)\right]=0$.
I hope this helps!
A: For the Lagrange's mean value theorem, there exists $\xi$ such that $n^2 < \xi < (n+1)^2$ and
$$f((n+1)^2) - f(n^2) = f'(\xi)(2n + 1) < \frac {2n + 1} {\xi^2} < \frac {2n + 1} {n^4}$$
