Uniform convergence of $f_n = (n^a x^2)/(n^2 +x^3)$

Question

My question is, if you have the sequence $$f_n = \frac{n^\alpha x^2}{n^2 +x^3}$$ on $[0, \infty)$, for values of a for $0<\alpha<2$ does the sequence uniformly converge?

I guess another way to think about it is what values can $a$ take such that

$$\lim_{n \rightarrow \infty} \left(\sup_{x \in [0,\infty)}\left[\frac{n^a x^2}{n^2 +x^3}\right]\right) = 0. $$

I'm having trouble proving this.. but I think that intuitively, $\alpha$ can be less than or equal to 1.

Thanks for the help.

Try calculating the maximum of $f_n(x)$ explicitly. – Antonio Vargas Nov 11 '13 at 03:51 — Antonio Vargas, Nov 11 '13 at 03:51
Here is a technique. – Mhenni Benghorbal Nov 11 '13 at 03:54 — Mhenni Benghorbal, Nov 11 '13 at 03:54

score 5 · Answer 1 · edited Apr 13 '17 at 12:20

5

If you work the problem out using the technique, then you will find that the maximum is achieved at

$$x=(2n^2)^{1/3}$$

and is given by

$$ \frac{ 2^{ \frac{2}{3}} }{3} n^{ a - \frac{2}{3}}.$$

Now, you should be able to finish the problem.

edited Apr 13 '17 at 12:20

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answered Nov 11 '13 at 04:14

Mhenni Benghorbal

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Uniform convergence of $f_n = (n^a x^2)/(n^2 +x^3)$

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