# Does this also imply uniform convergence? and hence preserves continuity on the interval $[0, \infty)$

I am asked to show pointwise convergence on $$[0, \infty)$$ of the series:

$$\sum_{n=1}^\infty \frac{1}{n^2 +n^3 x}$$

But I could just remark that since $$x \geq 0$$ we have that $$n^2 +n^3 x> n^2$$ $$\left|\frac{1}{n^2 +n^3 x}\right|=\frac{1}{n^2 +n^3 x} \leq \frac{1}{n^2} =M_n$$ Since $$|f_n| \leq M_n$$ and $$\sum M_n$$ convergent p-series, by the Weierstrass $$M$$ test, the original series converges uniformly on the domain.

Now I simply conclude that uniform convergence implies pointwise convergence.

The reason I want to show this is that in the next question we are asked to show that:

Prove that the function $$s: [0, \infty) \rightarrow \mathbb R$$ defined by the series, is continuous.

We know that uniform convergence preserves continuity. We only need to make the case that the individual functions are continuous. We know that quotients of polynomials are continuous, as long as we do not divide by zero, so $$\frac{1}{n^2+n^3 x}$$ runs into a problem if $$x=-\frac{1}{n}$$, but $$n>0$$ and $$x\geq0$$ so this can never happen. Therefore our polynomial quotient is continuous and so is the original function by uniform continuity.

• So yes, even though the question only asked for pointwise, if I prove uniform convergence, I can get continuity of the function series. – Wesley Strik Jan 27 '19 at 9:35

Yes, you are correct the function $$f(x)=\sum_{n=1}^\infty \frac{1}{n^2 +n^3 x}$$ is continuous in $$[0,+\infty)$$ because it is the uniform limit of the sequence of continuous functions, $$f_N(x)=\sum_{n=1}^N \frac{1}{n^2 +n^3 x}.$$ Indeed, as $$N$$ goes to infinity, $$\sup_{[0,+\infty)}|f(x)-f_N(x)|\leq \sum_{n=N+1}^\infty \frac{1}{n^2}\to 0$$ because $$\sum_{n=1}^\infty \frac{1}{n^2}$$ is a convergent series.

You've used the Weierstrass M test incorrectly. A series $$\sum f_n$$ of positive functions that satisfies

$$\sum_{n=1}^{\infty}f_n(x) \le \sum_{n=1}^{\infty}\frac{1}{n^2}$$

need not be uniformly convergent. However, what we have is

$$0\le \frac{1}{n^2+n^3x} \le \frac{1}{n^2}$$

for $$n=1,2,\dots$$ and $$x\ge 0.$$ Now Weierstrass M gives the desired result.

• "the original series with only positive terms " – Wesley Strik Jan 27 '19 at 20:18
• You might want to read my answer again. I was assuming the $f_n$ are positive. W-M requires an estimate on each $f_n,$ not on $\sum f_n.$ – zhw. Jan 27 '19 at 21:31
• the $f_n$ are the terms – Wesley Strik Jan 27 '19 at 21:43
• What you wrote was $$\sum_{n=1}^\infty \frac{1}{n^2 +n^3 x} \leq\sum_{n=1}^\infty \frac{1}{n^2}$$ What I'm saying is you can't apply WM based on that. i don't know how to make it clearer. – zhw. Jan 27 '19 at 22:09