Prove uniform convergence of $\sum\limits_{n=1}^\infty \frac{-1}{n(nx+1)^2}$ on $[a,\infty)$

Question

I need to prove that the limit function of a function series is differentiable on $[a,\infty)$, where $a>0$. I wanted to use the theorem that the function series has a derivative if the function is continuously differentiable, the function series is pointwise convergent and and the function series with the derivative of the function is uniformly convergent. The function is
$\sum\limits_{n=1}^\infty \frac{1}{n^2+n^3x}.$
I have proven that it is continuous and uniformly convergent, hence also pointwise convergent.
In order to prove that
$\sum\limits_{n=1}^\infty \frac{-1}{n(nx+1)^2}$ is uniformly convergent on $[a,\infty)$, I wanted to prove that $\sum\limits_{n=1}^\infty \frac{-1}{n(na+1)^2}$ is pointwise convergent and then use the Weierstrass M-test to conclude that $\sum\limits_{n=1}^\infty \frac{-1}{n(nx+1)^2}$ is uniformly convergent on $[a,\infty)$. However I can prove that it is uniformly convergent for $a>1$ and $a=1$, as $\frac{1}{n^2}$ is a convergent majorant, but I got stuck if $0<a<1$. So how could I prove that it is then also convergent? The ratio test fails.

Answer 1

Using the notation of the Wikipedia article on the Weierstrass M test: You want to know if $\sum_{n\ge1} \frac{-1}{n(nx+1)^2}$ converges uniformly on $A=[a,\infty)$,where $a$ is a positive real constant. Note that for $x\in A$ you have $|f_n(x)|\le \frac 1 {n(na+1)^2}$. But $$\sum_{n\ge1} M_n \le \frac 1 {a^2} \sum_{n\ge1}\frac 1 {n^3}<\infty,$$ which you should recognize as a convergent series.