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I know how to show that the following series will converge absolutely. But am unsure how to show it will or will not converge uniformly for $z\in (0,1).$

$\displaystyle \sum_{n \mathop = 1}^{\infty} \left( {\frac 1 {(z+n)^{1/2}}}-{\frac 1 {n^{1/2}}}\right)$

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  • $\begingroup$ What is the domain for $z$? $\endgroup$
    – Clement C.
    Jul 28 '15 at 13:35
  • $\begingroup$ Sorry, z is Real, and in the interval form (0,1) $\endgroup$ Jul 28 '15 at 13:49
  • $\begingroup$ Well i used this other article found on stack exchange math.stackexchange.com/questions/221994/… ... I'll post that instead of rewriting it. He's showing it for Dirichlet Eta but its generally the same for what i need. $\endgroup$ Jul 28 '15 at 14:01
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You have

$$\frac{1}{(z+n)^{1/2}}-\frac{1}{n^{1/2}}=\frac{n^{1/2} - (z+n)^{1/2}}{n^{1/2} (z+n)^{1/2}}$$

By the mean value theorem, $(z+n)^{1/2}=n^{1/2} + \frac{\xi_n}{2 n^{1/2}}$, where $\xi_n \in (0,1)$. Also $(z+n)^{1/2} \geq n^{1/2}$. Putting things together, the numerator is at most $\frac{1}{2 n^{1/2}}$ in magnitude while the denominator is at least $n$. Can you finish now?

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  • $\begingroup$ lol yes i think so, thanks for your help $\endgroup$ Jul 28 '15 at 14:08
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We have $$0>\sum_{n\geq1}\left(\frac{1}{\left(z+n\right)^{1/2}}-\frac{1}{n^{1/2}}\right)=-\frac{1}{2}\sum_{n\geq1}\int_{n}^{n+z}\frac{1}{x^{3/2}}dx\geq-\frac{1}{2}\sum_{n\geq1}\int_{n}^{n+1}\frac{1}{x^{3/2}}dx>$$ $$-\frac{1}{2}\sum_{n\geq1}\frac{1}{n^{3/2}}>-\infty. $$ If you prefer, we can use directly the M- test with the same passages $$\left|\frac{1}{\left(z+n\right)^{1/2}}-\frac{1}{n^{1/2}}\right|=\frac{1}{n^{1/2}}-\frac{1}{\left(z+n\right)^{1/2}}=\frac{1}{2}\int_{n}^{n+z}\frac{1}{x^{3/2}}dx\leq\frac{1}{2}\frac{1}{n^{3/2}}. $$

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  • $\begingroup$ -.5 Zeta(1.5) > - (infinity)? $\endgroup$ Jul 28 '15 at 14:21
  • $\begingroup$ $\zeta(1.5)< \infty $? If you prefer, we can show it in the positive situation, it's the same. $\endgroup$ Jul 28 '15 at 14:24
  • $\begingroup$ Im so sorry i read the inequalities incorrectly. I apologize. $\endgroup$ Jul 28 '15 at 14:30
  • $\begingroup$ @T.Poindexter No problem! ;) $\endgroup$ Jul 28 '15 at 14:32
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    $\begingroup$ @ManeeshNarayanan Right. $\endgroup$ Oct 23 '17 at 16:36
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Hint: showing normal convergence is usually easier, and a sufficient condition for uniform convergence. (when and if the test for normal convergence fails, then it is time to start thinking about the bad possible situation — that there may be uniform convergence nevertheless.)

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  • $\begingroup$ I think this would be more helpful if $z$ were complex, but it is evidently real (and positive). $\endgroup$
    – Ian
    Jul 28 '15 at 14:01
  • $\begingroup$ @Ian -- the method still can apply: the $\sup$ termwise is attained for $z=1$, and the corresponding series does converge. $\endgroup$
    – Clement C.
    Jul 28 '15 at 14:02
  • $\begingroup$ Fair point, though then the estimation from there is the same as the estimation I did anyway :) $\endgroup$
    – Ian
    Jul 28 '15 at 14:04
  • $\begingroup$ @Ian True :) ${}$ $\endgroup$
    – Clement C.
    Jul 28 '15 at 14:05
  • $\begingroup$ Well generally i would have liked z to be an element of the complex open unit disk with deleted origin, but really its immaterial as Z as a real variable was sufficient for what i needed $\endgroup$ Jul 28 '15 at 14:15

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