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

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. So how could I prove that it is then also convergent? The ratio test fails.

• Can you use the fact that $an\le an+1$? – kimchi lover Jan 25 at 16:58
• I don't think so, because then you get $n(na)^2=n^3a$, but you can't be sure that this is smaller than $n^2$ – C. Elias Jan 25 at 17:40

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.