for a differentiable function $f$ in $(0,\infty)$ and 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.
for a differentiable function $f$ in $(0,\infty)$ and 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.
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}
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}.
The mean value theorem can serve you here.
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, there exists $c$ such that $a
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
(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!