Does the following series converge uniformly?

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 $z\in (0,1).$

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

• What is the domain for $z$? Jul 28 '15 at 13:35
• Sorry, z is Real, and in the interval form (0,1) Jul 28 '15 at 13:49
• 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. Jul 28 '15 at 14:01

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?

• lol yes i think so, thanks for your help Jul 28 '15 at 14:08

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 $$\left|\frac{1}{\left(z+n\right)^{1/2}}-\frac{1}{n^{1/2}}\right|=\frac{1}{n^{1/2}}-\frac{1}{\left(z+n\right)^{1/2}}=\frac{1}{2}\int_{n}^{n+z}\frac{1}{x^{3/2}}dx\leq\frac{1}{2}\frac{1}{n^{3/2}}.$$

• -.5 Zeta(1.5) > - (infinity)? Jul 28 '15 at 14:21
• $\zeta(1.5)< \infty$? If you prefer, we can show it in the positive situation, it's the same. Jul 28 '15 at 14:24
• Im so sorry i read the inequalities incorrectly. I apologize. Jul 28 '15 at 14:30
• @T.Poindexter No problem! ;) Jul 28 '15 at 14:32
• @ManeeshNarayanan Right. Oct 23 '17 at 16:36

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.)

• I think this would be more helpful if $z$ were complex, but it is evidently real (and positive).
– Ian
Jul 28 '15 at 14:01
• @Ian -- the method still can apply: the $\sup$ termwise is attained for $z=1$, and the corresponding series does converge. Jul 28 '15 at 14:02
• Fair point, though then the estimation from there is the same as the estimation I did anyway :)
– Ian
Jul 28 '15 at 14:04
• @Ian True :) ${}$ Jul 28 '15 at 14:05
• 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 Jul 28 '15 at 14:15