An application of Weierstrass M-Test to determine sets of uniform convergence.

Question

let $\sum_{n=1}^{\infty}fn(x) $ = $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}(\frac{x}{1+x})^n$
I am attempting to apply Weierstrass M-Test on [a,$\infty$)
I get that
$|\frac{1}{\sqrt{n}}(\frac{x}{1+x})^n|$<$|\frac{1}{\sqrt{n}}|$
but the RHS doesnt converge, so we cant use M-Test.
What other methods can I use?

Answer 1

You cannot apply on $[a,+\infty)$, because $$M_n:=\sup_{x\ge a} |f_n(x)|=\lim_{x\to+\infty} f_n(x)=\frac{1}{\sqrt n}.$$ So you this is teh best bound and Weierstrasss M-Test cannot be applied.

But you can use it in $[a,b]$ for all $b>0$, since the series of genreral term $\frac{1}{\sqrt n} \left(\frac{b}{b+1}\right)^n$ is convergent.

With this method, you can prove that this series is uniform convergent on any compact set of $[a,+\infty)$.

Answer 2

Let $g_m(x)=\sum_{n=1}^m \frac {1}{\sqrt n}\left(\frac {1}{1+x}\right)^n.$

If $S$ is a subset of $\Bbb R$ with no upper bound in $\Bbb R$ then $g_n$ does not converge uniformly on $S.$

Proof: We show that for any $m\in \Bbb N$ there exist $m'>m$ and $x\in S$ with $g_{m'}(x)-g_m(x)>1/(2\sqrt 2),$ as follows:

Let $m'=2m.$ Now take $x\in S$ where $x>0$ and $x$ is large enough that $(x/(1+x))^{2m}>1/2.$ Then $(x/(1+x))^n>1/2$ when $m<n\le 2m,$ so $$g_{m'}(x)-g_m(x)=g_{2m}(x)-g_n(x)=\sum_{n=m+1}^{2m}(1/\sqrt n)(x/(1+x))^n>$$ $$>\sum_{n=m+1}^{2m}(1/\sqrt n)(1/2)\ge$$ $$\ge \sum_{n=m+1}^{2m}(1/\sqrt {2m})(1/2)=$$ $$=m(1/\sqrt {2m})(1/2)=(\sqrt m)/(2\sqrt 2)\ge 1/(2\sqrt 2).$$