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How to show that $$\sum_{n=1}^{\infty}\frac{x^{2n}}{(x+n)^{3/2}}$$ is uniformly continuous on $[0,1]$

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Hint: Show that $\frac{x^{2n}}{(x+n)^{3/2}}\leq \frac 1{n^{3/2}}$, hence the series is _ (fill in the blank) convergent. What about the sum of a series of uniformly continuous functions which is __ (fill in the blank) convergent?

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    The title and question don't seem to match :-)2012-04-01
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    @Aryabhata I agree.2012-04-01
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    Anyway, here the series converges uniformly (Weistrass M-test similar to what you noted) and so the limit function is continuous, and since the interval is $[0,1]$, the function is uniformly continuous too. Not too familiar with normal convergence and I suppose it is possibly out of scope of a beginning calculus course.2012-04-01
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    Yes, in fact the dots can be also for uniformly.2012-04-01
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    On an intial reading, those dots seemed to indicate a pause for dramatic effect :-) and not something to be filled in!2012-04-01
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    You are right, I don't know how to do it in a more convenient way.2012-04-01
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    which is _________ convergent (fill in the blank)? (Using underscore character...)2012-04-01
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    @Aryabhata And all this time I was using "\underline{\phantom{Iwantittobethiswide}}"2012-04-01
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    ............. :-)2012-04-01
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WolframAlpha says:

$\sum_{n=1}^{\infty}\frac{x^{2n}}{(x+n)^{3/2}}= x^2 \Phi\left( x^2,\frac{3}{2},x+1 \right)$, when $|x|<1$. $\Phi(z,s,a)$ gives the Hurwitz Lerch transcendent.