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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.2012-05-31
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    yes... convergent. Sorry!2012-05-31
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    No problem. Did you try the ratio test?2012-05-31
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    Yes. I just tried it, but couldn't reach the conclusion...2012-05-31

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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!2012-05-31
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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?2012-05-31
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    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).2012-05-31