the uniform convergence of series (M-test).

Question

$\sum_{n=1}^{\infty} \dfrac{\arctan n + \sqrt {|x|}}{n^2}$. Does the series converge uniformly on $R$. ?

Let $f_k(x) = \dfrac{\arctan k + \sqrt {|x|}}{k^2}$. Then, I have to find $M_k$ such that $|f_k(x)| \le M_k$ for all $x \in R$. Concurrently, $\sum_{k=1}^{\infty} M_k$ has to converge. I think such $M_k$ does not exist because $\sqrt{|x|}$ can approach to infinity.

If I am correct, I have to disprove this, but I am struggling to find a way to disprove it. Could you give some hint?

Thank you in advance.

Answer 1

It's not uniformly Cauchy: Fix $N\in \mathbb{N}$, then for any $k\geq 1$ we ought to be able to keep $$ \sup_{x\in \mathbb{R}}\left|\sum_{n=N}^{N+k} \frac{\arctan(n)+\sqrt{|x|}}{n^2}\right| $$ small by making $N$ large. However, $$ \sup_{x\in \mathbb{R}}\left|\sum_{n=N}^{N+k} \frac{\arctan(n)+\sqrt{|x|}}{n^2}\right|= \sup_{x\in \mathbb{R}}\sum_{n=N}^{N+k} \frac{\arctan(n)+\sqrt{|x|}}{n^2}\\ \geq \sup_{x\in\mathbb{R}} \sum_{n=N}^{N+k} \frac{\sqrt{|x|}}{n^2}\\ \stackrel{|x|=(N+k)^4}{\geq} (N+k)^2\sum_{n=N}^{N+k} \frac{1}{n^2}\\ \geq (N+k)^2 k\left(\frac{1}{(N+k)^2}\right)\\ =k\not\to0 $$ as $N\to \infty$.

Answer 2

Let

$$R_n(x)=\sum_{k=n}^{\infty}\dfrac{\arctan k+\sqrt{|x|}}{k^2}$$

This is the rest of the series.

Uniform convergence would mean that

$$\forall \varepsilon>0,\exists N,\forall n>N,\forall x\in\Bbb R, |R_n(x)|<\varepsilon$$

If you suspect that the series does not converge uniformly, then you have to prove the contrary:

$$\exists \varepsilon>0,\forall N,\exists n>N, \exists x\in\Bbb R, |R_n(x)|\ge\varepsilon$$

Now, let $n\ge1$, you have

$$R_n(x)>\sqrt{|x|}\sum_{k=n}^\infty\dfrac{1}{k(k+1)}=\frac{\sqrt{|x|}}{n}$$

It should not be difficult to conclude.