Does $\sum_{k=1}^{\infty}\frac{2^{k-1} x^{2^{k-1}-1}}{1+x^{2^{k-1}}}$ converges uniformly. $-1<x<1$

I have tried to bound it by Weierstrass M-Test but haven't been successful. I have also tried to prove that the sequence of partial sums is Cauchy but haven't managed to do that. Please suggest on how to do this.

This is in relation to another question which I asked yesterday, the link is given below. (Proving pointwise convergence of series of functions).

In response to that user "Science" came up with something which essentially needs to show that this series is uniformly convergent.

  • $\begingroup$ Are you looking for uniform convergence the domain $\mathbb{R}$ ? $\endgroup$ – Srinivas K Feb 20 '15 at 4:00
  • $\begingroup$ Sorry domain is (-1,1), I should have mentioned $\endgroup$ – Silver moon Feb 20 '15 at 4:03
  • $\begingroup$ Maybe we could try Abel's test for uniform convergence. Have you tried this already ? $\endgroup$ – Srinivas K Feb 20 '15 at 4:39
  • $\begingroup$ Sorry I am not aware of this test. Can you please do this for me? Thanks $\endgroup$ – Silver moon Feb 20 '15 at 4:43

The series does not converge uniformly on $(0,1)$ and hence also not on $(-1,1)$.

To see this, let $f_n (x) = \frac{2^{n-1} \cdot x^{2^{n-1} - 1}}{1 + x^{2^{n-1}}}$. Then each $f_n$ is continuous and (hence) bounded on $[0,1]$.

Now, let $f(x) := \sum_n f_n(x)$. If $\sum_{n=1}^N f_n \to f$ would hold with uniform convergence on $(0,1)$, it would follow that $f$ is bounded on $(0,1)$.

But since each term of the sum is nonnegative for $x \in (0,1)$, we have

$$ \Vert f \Vert_\sup \geq f(x) \geq f_n(x) \geq \frac{2^{n-1} x^{2^{n-1}-1}}{1 + x^{2^{n - 1}}} \to \frac{2^{n-1}}{2} \text{ for } x \uparrow 1. $$

Since this holds for arbitrary $n \in \Bbb{N}$, we conclude $\Vert f \Vert_\sup = \infty$.

Hence, $f$ is not bounded on $(0,1)$, so that the series is not uniformly convergent on $(0,1)$.

  • $\begingroup$ ok but can you please check this link math.stackexchange.com/questions/1155407/… $\endgroup$ – Silver moon Feb 20 '15 at 7:53
  • $\begingroup$ In the answer by user "science", how is he justified to interchange the limits in integration when the series doesn't converge uniformly? $\endgroup$ – Silver moon Feb 20 '15 at 7:54
  • $\begingroup$ @Silver moon: One possibility is to consider the series only on $[-\alpha, \alpha]$ for fixed (but arbitrary) $0 < \alpha < 1$. There, the series does converge uniformly. Afterwards, you can deduce pointwise convergence on $(-1,1)$, since $ 0 < \alpha <1 $ was arbitrary. $\endgroup$ – PhoemueX Feb 20 '15 at 8:13
  • $\begingroup$ Can you elaborate, this has been my problem. I have been trying to prove its uniform convergence on closed sub-interval but haven't been able to bound this $\endgroup$ – Silver moon Feb 20 '15 at 8:19
  • $\begingroup$ @Silver moon: Note that for $|x| \leq \alpha < 1$, we have $|x^{2^{n - 1}}| \leq \alpha^{2^{n - 1}} \leq \alpha$ (for $n>1$) and hence $1 + x^{2^{n-1}} \geq 1 - \alpha$. Hence, you can estimate each term by $| \frac{2^{n-1} x^{2^{n-1}-1}}{1 + x^{2^{n-1}}}| \leq \frac{2^{n-1} x^{2^{n-1}-1}}{1-\alpha}$. No verify that $\sum_n n \cdot x^n$ has radius of convergence $1$. This should allow you to conclude the proof. $\endgroup$ – PhoemueX Feb 20 '15 at 8:38

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