A strange (seemingly pointless) exercise on convergence of series

I have come across an exercise which asks to prove that the series of functions$$\sum\frac{x^n}{1+x^n}$$ is convergent for $x\in [0,1)$. It also asks us to prove that the series converges uniformly for each $a: 0<a<1$. I think I know how to answer the second question (proving the uniform convergence of the series): $$\sum\frac{x^n}{1+x^n}\leq\sum x^n$$ We can then use the Weierstrass M-test and let $\sum M_k$= $\sum x^n$, which then converges to $\frac{1}{1-x}$ for $\vert x\vert<1$. Hence $\sum\frac{x^n}{1+x^n}$ is uniformly convergent. However, I am not sure how to answer the first question. I think that I could just say that since the series is uniformly convergent, it must converge. Or I could use the same inequality as above and use the comparison test to show that it converges. However, I am not sure. What am I supposed to do? Is it asking whether the series of functions is pointwise convergent? If not what is the difference between proving that a series of functions converges pointwise vs. proving that a series of functions converges? Am I missing something? Also, is my approach in proving uniform convergence correct?

• What series of functions? Where is $a$ used? Did you mean to have $x^n$ instead of $x_n$? – copper.hat Nov 17 '13 at 8:29
• Yes, sorry. It should be a superscript. – CoffeeIsLife Nov 17 '13 at 8:37
• Also, $a$ should be the value of $x$ – CoffeeIsLife Nov 17 '13 at 8:48
• Did you leave something out of the problem statement? Were you asked to prove uniform convergence on the set [0,a] for each $a$? Is that what you proved, or did you prove more, that the series converges uniformly on the set $[0,1)$? – bof Nov 17 '13 at 9:13
• @No, it’s not the case. But there is no difficulty indeed, in some way it is simpler to answer the second question before the first. What happens is this : the series converges uniformly on every $[0,a]$, so it converges simply on every $[0,a]$, so it must converge simply on the union of all those $[0,a]$'s, which is $[0,1)$. But it is not true that the series converges uniformly on $[0,1)$. If it did, it would converge at $x=1$ also. – Ewan Delanoy Nov 17 '13 at 12:04

We can then use the Weierstrass $M$-test and let $\sum M_k= \sum x^n$
No. The whole point of $M$-test is that $M$-numbers do not involve $x$. What should be done instead: on the interval $[0,a]$ we have $$\frac{x^n}{1+x^n}\le x^n \le a^n$$ Let $M_n=a^n$ and apply the test.
Yes. More precisely: every point $x\in [0,1)$ is contained in some interval $[0,a]$ with $a<1$ (take any $a$ with $x<a<1$), and we already know that the series converges, even uniformly, on $[0,a]$. Thus it converges at $x$; and this is all we need to show to demonstrate pointwise convergence on $[0,1)$.
Apparently, you were not asked to show that uniform convergence fails on $[0,1)$. For the same of completeness, here is a sketch: for every $n$ there is $x\in [0,1)$ such that $x^n=1/2$. Hence $\frac{x^n}{1+x^n}=1/3$. So, the terms of the series do not tend to $0$ uniformly, thus failing a necessary condition for uniform convergence.
• Nice :). We can also prove the uniform convergence by proving the normal convergence on $[0,a]$ – Nour May 17 '17 at 14:05