For what $\alpha$ is the series uniformly convergent on $[0,\infty)$

Question

The problem is stated as:

Prove that $\sum_{n=1}^{\infty} \frac{x^\alpha}{\sqrt{n} (n^2+x^3)}$ is uniformly convergent on $[0, \infty)$ for $\alpha = 2$.

I thought I might challenge myself a bit, so from hereon, I'll try to show for what $\alpha$ in general, that this series converges uniformly on $[0,\infty)$.

My attempt

Let $u_n(x) := \frac{x^\alpha}{\sqrt{n} (n^2+x^3)}$. We want to use Weierstrass M Test, in order to do this, we want to find $\sup_{x\in[0,\infty)} u_n(x)$ Observe that $\alpha \geq 0$ in order to have continuity at $ x=0$. Let's work with $\alpha \geq 0$ from hereon.

Taking the derivative, we get: $u'_n(x) = \frac{x^{\alpha - 1}\sqrt{n}[\alpha n^2+(\alpha-3)x^3]}{\sqrt{n}(n^2+x^3)^2}$

Obseve that $u_n(x)$ becomes a decreasing function for a certain x - value, let's call it $\hat{x}$ if we let $\alpha < 3$. For $\alpha \geq 3$ we have that $u_n(x)$ are monotonically increasing as functions. In that case, we can't find a supremum of our $u_n(x)$'s, where we wouldn't be able to apply Weierstrass M Test.

Solving for $x$ in the case that $\alpha < 3$, we get that we have a maximum at $\hat{x} = (\frac{\alpha}{3-\alpha}n^2)^{1/3}$

Hence, we have that: $M_n(x) := \sup_{x\in[0,\infty)} |u_n(x)| = |u_n(\hat{x})| = |\frac{(\frac{\alpha}{3-\alpha}n^2)^{1/3})^\alpha}{\sqrt{n} (n^2+(\frac{\alpha}{3-\alpha}n^2))}|$ which asymptotically behaves as $n^{2\alpha/3-5/2}$ and for which we set $2\alpha/3-5/2<-1$ in order to apply Weierstrass M test. Thus, we get that $\alpha < 9/4$, and with what we find above, we decude that we have uniform convergence for $\alpha \in [0, 9/4)$

I'd be glad if you could share any tips if you find anything wrong in the solution.

Thanks!

Answer 1

I think OP approach is good but here is a somewhat more complete analysis.

Note that the sum always converges since it can be bounded by $\leq c n^{-2}$ for some $c$ (depending on $x$).

One way to tackle the problem of uniform convergence is to consider the largest value per summand.

First, we can easily discard $\alpha \geq 3.$ Indeed, uniform convergence signifies that the sequence $\gamma_p = \sup\limits_{x \in [0, \infty)} \sum\limits_{n \geq p} \dfrac{x^\alpha}{\sqrt{n}(n^2 + x^3)}$ obeys $\gamma_p \to 0.$ Clearly $\gamma_p \geq \sup\limits_{x \in [0, \infty)} \dfrac{x^\alpha}{\sqrt{p} (p^2 + x^3)}.$ If $\alpha = 3,$ the right hand side equals 1 and if $\alpha > 3$ the right hand side is unbounded. Therefore, $\alpha < 3$.

Consider $u = u_\beta = \dfrac{x^\alpha}{\beta + x^3}.$ Then $u$ is maximised at $x = \sqrt[3]{\dfrac{\alpha \beta}{3 - \alpha}}.$ And considering anything but $\beta$ a constant, we see that $u \leq \hat{u} := u\left( \sqrt[3]{\dfrac{\alpha \beta}{3 - \alpha}}\right) \asymp \beta^{-(1 - \frac{\alpha}{3})}$ where $X \asymp Y$ means that the $aX \leq Y \leq b X$ for some "universal constants" $a$ and $b$ (the constants depend on $\alpha$ but not on $\beta$). Then, $$ \sum\limits_{n = 1}^\infty \dfrac{x^\alpha}{\sqrt{n}(n^2 + x^3)} \leq \sum_{n = 1}^\infty \dfrac{\hat{u}_{n^2}}{\sqrt{n}} \asymp \sum_{n = 1}^\infty \dfrac{n^{-(2 - \frac{2\alpha}{3})}}{\sqrt{n}} = \sum_{n = 1}^\infty n^{-\zeta} $$ where $\zeta = 2 + \dfrac{1}{2} - \dfrac{2\alpha}{3}.$ The right hand side converges when $\zeta > 1$ which entails $\alpha < \dfrac{9}{4}.$ Therefore, there there is uniform convergence for $0 \leq \alpha < \dfrac{9}{4}.$

The question remains open for $\dfrac{9}{4} \leq \alpha < 3.$ However, I suspect there will not be uniform convergence. Note that the proof gives that the maximums of each summand are at $c n^{\frac{2}{3}}$ for some $c$ depending solely on $\alpha.$ This says that the tail $\gamma_p$ will always take into account the worst-case scenario for an infinite number of summands, and the function $x \mapsto \dfrac{x^\alpha}{\beta + x^3}$ is quite flat for $\alpha < 3$ but close to 3.

Answer 2

I believe OP already showed, using the Weierstrass M-test, that the series converges uniformly in $x \in [0, \infty)$ if $\alpha < \frac{9}{4}$. So, I will show that the series does not converge uniformly if $\alpha \geq \frac{9}{4}$.

Note that the series converges uniformly in $x \in [0, \infty)$ if and only if the tail series converges to $0$ uniformly in $x \in [0, \infty)$:

$$ \bbox[border:1px blue;padding:7px;background-color:azure;]{\lim_{N\to\infty} \sup_{x\geq 0} \Biggl( \sum_{n=N}^{\infty} \frac{x^\alpha}{\sqrt{n} (n^2+x^3)} \Biggr) = 0.} $$

Now suppose $\alpha \geq \frac{9}{4}$. Then by noting that $n \mapsto \frac{x^{9/4}}{\sqrt{n} (n^2+x^3)}$ is decreasing in $n$, for $x \geq 1$ we get

$$ \int_{N}^{\infty} \frac{x^{9/4}}{\sqrt{s}(s^2+x^3)} \, \mathrm{d}s \leq \sum_{n=N}^{\infty} \frac{x^{9/4}}{\sqrt{n} (n^2+x^3)} \leq \sum_{n=N}^{\infty} \frac{x^\alpha}{\sqrt{n} (n^2+x^3)}. $$

However, by substituting $s = x^{3/2}t$,

$$ \int_{N}^{\infty} \frac{x^{9/4}}{\sqrt{s}(s^2+x^3)} \, \mathrm{d}s = \int_{N/x^{3/2}}^{\infty} \frac{1}{\sqrt{t}(t^2+1)} \, \mathrm{d}t \to C \qquad \text{as } x \to \infty, $$

where $C = \int_{0}^{\infty} \frac{\mathrm{d}t}{\sqrt{t}(t^2+1)} = \frac{\pi}{\sqrt{2}}$. Combining altogether, it follows that

$$ \sup_{x\geq 0} \Biggl( \sum_{n=N}^{\infty} \frac{x^\alpha}{\sqrt{n} (n^2+x^3)} \Biggr) \geq \lim_{x\to\infty} \sum_{n=N}^{\infty} \frac{x^\alpha}{\sqrt{n} (n^2+x^3)} \geq C $$

for any $N$ and therefore the series does not converge uniformly in $[0,\infty)$ for $\alpha \geq \frac{9}{4}$.