# Prove that a series is not uniformly convergent

How can i show that the series

$$\sum_{n=1}^{\infty}\frac{x^{2n}}{(x+n)^{3/2}}$$ is not uniformly convergent on $[1,\infty)$

I know that for given $\epsilon > 0$, we need to construct a sequence $x_n$ such that $|\frac{x^{2n}}{(x+n)^{3/2}}| > \epsilon$. The thing is I was not able to come up with such a sequence.

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Do you mean "convergent" rather than "continuous"? I recommend using the ratio test to check for convergence. – Jonas Meyer May 31 '12 at 19:45
yes... convergent. Sorry! – user9292 May 31 '12 at 19:56
No problem. Did you try the ratio test? – Jonas Meyer May 31 '12 at 19:56
Yes. I just tried it, but couldn't reach the conclusion... – user9292 May 31 '12 at 19:59

It is not even convergent when $x>1$. Some ways to see this:

• Ratio test: $$\lim_{n\to\infty}x^2\left(\frac{x+n}{x+n+1}\right)^{3/2}=x^2.$$
• Root test: $$\lim_{n\to\infty}\frac{x^2}{(x+n)^{3/(2n)}}=x^2.$$
• Divergence test: $$\lim_{n\to\infty}\frac{x^{2n}}{(x+n)^{3/2}}$$ can be calculated with two applications of l'Hôpital's rule if $x>1$.
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That's wonderful. Now I got it. Many thanks! – user9292 May 31 '12 at 20:13

Hint : given $n \in \mathbb{N}$, how does the function $\frac{x^{2n}}{(x+n)^{3/2}}$ behave when $x$ goes to $+\infty$ ?

You should be able to complete the proof after having answered that question.

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indeed, it approaches infinite. So showing this would be sufficient as a proof? – user9292 May 31 '12 at 20:06
with your idea of proof, if it approaches infinity, then a fortiori there is a $x_n$ such that it is greater than $\epsilon$ (as noted by Jonas, it diverges rather crudely). – Glougloubarbaki May 31 '12 at 21:28