Does the following series converge uniformly?

Question

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 $z\in (0,1).$

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

Well i used this other article found on stack exchange http://math.stackexchange.com/questions/221994/riemann-zeta-function-manipulation/222588#222588 ... I'll post that instead of rewriting it. He's showing it for Dirichlet Eta but its generally the same for what i need. — T. Poindexter, Jul 28 '15 at 14:01

score 1 · Answer 1 · answered Jul 28 '15 at 13:58

1

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?

answered Jul 28 '15 at 13:58

Ian

101,645

lol yes i think so, thanks for your help – T. Poindexter Jul 28 '15 at 14:08

Marco Cantarini · Answer 2 · 2015-07-28T14:32:04.380

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

edited Jul 28 '15 at 14:32

answered Jul 28 '15 at 14:13

Marco Cantarini

33,062
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-.5 Zeta(1.5) > - (infinity)? – T. Poindexter 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. – Marco Cantarini Jul 28 '15 at 14:24
Im so sorry i read the inequalities incorrectly. I apologize. – T. Poindexter Jul 28 '15 at 14:30
@T.Poindexter No problem! ;) – Marco Cantarini Jul 28 '15 at 14:32
@MarcoCantarini How do $-\frac{1}{2}$ $ \sum_{n =1}^{\infty}$$ \int_{n}^{n+z}$ $\frac{1}{x^{3/2}}dx$ $\ge$ $-\frac{1}{2}$ $\sum_{n=1}^{\infty}$ $ \int_{n}^{n+1}$ $\frac{1}{x^{3/2}}dx$ come? Please help me. – Oct 23 '17 at 06:17
@ManeeshNarayanan Note that $z \in (0,1)$ and the integrand is positive. – Marco Cantarini Oct 23 '17 at 14:25
@MarcoCantarini Then, [n,n+z] is smaller interval no than [n, n+1] then, negative sign is reverting the inequality, right? – Oct 23 '17 at 15:12
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@ManeeshNarayanan Right. – Marco Cantarini Oct 23 '17 at 16:36

Clement C. · Answer 3 · 2015-07-28T14:01:56.733

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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.)

edited Jul 28 '15 at 14:01

answered Jul 28 '15 at 14:00

Clement C.

67,323

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. – Clement C. 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 :) ${}$ – Clement C. 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 – T. Poindexter Jul 28 '15 at 14:15

Does the following series converge uniformly?

3 Answers3