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for a differentiable function $f$ in $(0,\infty)$ and 0 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.

  • 1
    So Right, Thank you so much.2011-06-22

4 Answers 4

0

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}

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}.

2

The mean value theorem can serve you here.

  • 1
    Note that in the case of $f(x)=\sqrt{x}$, where $f'(x) = \frac{1}{{2\sqrt x }}$, it holds $\frac{1}{{2\sqrt {(n+1)^2 } }} \leq f'(\xi) \leq \frac{1}{{2\sqrt {n^2 } }}$, that is $\frac{1}{{2(n+1)}} \leq f'(\xi) \leq \frac{1}{{2n}}$; hence $\frac{{2n + 1}}{{2(n + 1)}} \le f'(\xi )(2n + 1) \le \frac{{2n + 1}}{{2n}}$, and so, by the squeeze theorem, $\lim _{n \to \infty } f'(\xi )(2n + 1) = 1$.2011-06-22
2

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 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.

(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!