Using the Weierstrass M-test properly,

Question

I am slightly confused by the last part of my work here. For $x>0$ and for $\alpha > \frac{1}{2}$,

$$\left |\frac{x}{(1+nx^2)n^{\alpha}}\right |$$

$$=\left |\frac{1}{(\frac{1}{x}+nx)n^{\alpha}}\right |$$

$$\le\left |\frac{1}{(nx)n^{\alpha}}\right |$$

$$\le \frac{1}{xn^{1+\alpha}}$$

and now I claim that

$$\sum_{n=1}^{\infty}\frac{1}{xn^{1+\alpha}}<\infty$$

by the p-series / integral tests, and so

$$\sum\frac{x}{(1+nx^2)n^{\alpha}}$$ converges uniformly (to a continuous function of $x$ >0) by the Weierstrass M-test.

Is my claim correct? The $\large \frac{1}{x}$ factor in the summand confuses me a little bit; but, it is fixed, so I don't think it affects the convergence of the p-series.

Any comments are welcome.

Thanks,

Answer 1

Edited to point out that we actually have uniform convergence on all of $\mathbb R$, not just $(0,\infty)$.

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Observe that $$0 \leq (\sqrt{n}|x| - 1)^2 = nx^2 - 2\sqrt{n}|x| + 1$$ so $$1 + nx^2 \geq 2\sqrt{n}|x|$$ and therefore, $$\frac{|x|}{1 + nx^2} \leq \frac{1}{2\sqrt{n}}$$ Consequently, for all $x \in \mathbb R$ we have $$\left|\frac{x}{(1+nx^2)n^{\alpha}}\right| = \frac{|x|}{(1+nx^2)n^{\alpha}} \leq \frac{1}{2n^{\alpha + 1/2}}$$ Now, observe that the right hand side does not depend on $x$, and its sum converges since $\alpha > 1/2$. So you can conclude from the Weierstrass $M$-test that $$\sum_{n=1}^{\infty}\frac{x}{(1+nx^2)n^{\alpha}}$$ converges uniformly for all $x \in \mathbb{R}$. Among other things, this allows us to conclude that the limit function $$F(x) = \sum_{n=1}^{\infty}\frac{x}{(1+nx^2)n^{\alpha}}$$ is continuous everywhere.

Answer 2

You cannot keep any $x$ in the upper bound -- you are looking for a uniform upperbound on $f_n(x) = \frac{x}{n^\alpha(1+nx^2)}$, i.e. something of the form $$ \exists (M_n)_{n\geq 1} \text{ s.t. }\forall n\geq 1, \forall x > 0,\ \lvert f_n(x)\rvert \leq M_n; \text{ and } \sum_{n=1}^\infty M_n < \infty $$ Here, your $M_n$ depends on $x$: this is not legit. A way to solve your problem would be to see if, for every $n\geq 1$, $\lvert f_n\rvert$ can be maximized by some value $x_n >0$; and set $M_n = \lvert f_n(x_n)\lvert$.

For instance, studying the function $f_n$ (and differentiating it), it appears that $f_n=\lvert f_n\rvert$ is maximized at $x_n = \frac{1}{\sqrt{n}}$. If you take $M_n = \lvert f_n(\frac{1}{\sqrt{n}})\lvert$, you do get (by construction!) $\lvert f_n(x)\rvert \leq M_n$ for all $x>0$. But then, it only remains to check whether $\sum_{n=1}^\infty M_n < \infty$.

Answer 3

In a similar fashion to this answer, we see, using the Beta Function, that $$ \begin{align} \lim_{x\to0}\,x^{1-2\alpha}\sum_{n=1}^\infty\frac{x}{(1+nx^2)n^{\alpha}} &=\lim_{x\to0}\sum_{n=1}^\infty\frac{x^2}{(1+nx^2)n^{\alpha}x^{2\alpha}}\\ &=\int_0^\infty\frac{\mathrm{d}t}{(1+t)t^\alpha}\\[6pt] &=\Gamma(1-\alpha)\Gamma(\alpha)\\[9pt] &=\frac{\pi}{\sin(\pi\alpha)} \end{align} $$ and we can bound $$ \frac{\pi x^{2\alpha-1}}{\sin(\pi\alpha)}-\frac{x}{1-\alpha}\le \sum_{n=1}^\infty\frac{x}{(1+nx^2)n^{\alpha}}\le\frac{\pi x^{2\alpha-1}}{\sin(\pi\alpha)} $$