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 $z\in (0,1).$

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

Answer 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?

Answer 2

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.)

Answer 3

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