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for 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.

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Hint: how would you prove that a differentiable function on $(0,\infty)$ with bounded derivative is uniformly continuous on $(0,\infty)$? Note that the condition in your question $0<f'(x)<\frac{1}{x^2}$ (for all $x>0$) does not imply that the derivative of $f'$ is bounded on $(0,\infty)$ (since $\frac{1}{x^2}$ "blows up" if $x$ is close to $0$). On the other hand, it does imply that the derivative of $f$ is bounded on $(1,\infty)$. In any case, the uniform continuity of $f$ on $(1,\infty)$ does not really help since $(n+1)^2-n^2=2n+1\to \infty$ as $n\to\infty$. –  Amitesh Datta Jun 22 '11 at 11:53
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So Right, Thank you so much. –  user6163 Jun 22 '11 at 11:56
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4 Answers 4

up vote 2 down vote accepted

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!

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The mean value theorem can serve you here.

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Here's a simple example where $f'(x) \downarrow 0$ as $x \to \infty$ but $f((n + 1)^2 ) - f(n^2 ) = 1$ for all $n$. Let $f(x)=\sqrt{x}$, $x > 0$. Then $f'(x)=\frac{1}{{2\sqrt x }}$ and $f((n + 1)^2 ) - f(n^2 ) = (n+1) - n =1$. –  Shai Covo Jun 22 '11 at 12:15
    
Shai: If $\lim_{x\to \infty}f'(x)=0$ what could we say about requested limit now? Isn't it the same? –  user6163 Jun 22 '11 at 14:20
    
Nir: In the above example of $f(x)=\sqrt{x}$, $\lim _{x \to \infty } f'(x) = 0$ and $\lim _{n \to \infty } (f((n + 1)^2 ) - f(n^2 )) = \lim _{n \to \infty } 1 = 1$. This indicates that you cannot conclude much from the condition $\lim _{x \to \infty } f'(x) = 0$. –  Shai Covo Jun 22 '11 at 14:47
    
But by applying the squeeze theorem you get that $f'(x)$ tends to 0 anyway, so what's the difference? You say that without knowing the specific information that was given it's not able to answer this question like the way we did? –  user6163 Jun 22 '11 at 14:53
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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, $\lim _{n \to \infty } f'(\xi )(2n + 1) = 1$. –  Shai Covo Jun 22 '11 at 20:45
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For the Lagrange's mean value theorem, there exists $\xi$ such that $n^2 < \xi < (n+1)^2$ and $$f((n+1)^2) - f(n^2) = f'(\xi)(2n + 1) < \frac {2n + 1} {\xi^2} < \frac {2n + 1} {n^4}$$

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We get the result thanks to the inequalities $$0\leq \int_{n^2}^{(n+1)^2}f'(t)dt=f((n+1)^2)-f(n^2) \leq \int_{n^2}^{(n+1)^2}\frac 1{x^2} dx = \frac{-1}x\mid_{n^2}^{(n+1)^2}=\frac 1{n^2}-\frac 1{(n+1)^2}.$$

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