Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For $r<1$ define $F(r)=\sum_{n\in\mathbb N}(-1)^nr^{2^n}$. Does $F$ have a limit as $r\nearrow 1$?

share|cite|improve this question
Did you try to compute the series for fixed $r<1$? – Rasmus Nov 22 '11 at 8:19
In order for the community to better assist you, it is helpful if you provide what you have tried so far and indicate precisely where you are having difficulties. – Austin Mohr Nov 22 '11 at 8:19
I think it is not computable. Am I wrong? – bx1 Nov 22 '11 at 8:20
I simply don't know how to approach the problem. – bx1 Nov 22 '11 at 8:21
Actually, the answer is 'no'. The High-Indice Theorem tells us that whenever $0 \leq n_0 < n_1 < n_2 < \cdots$ satisfies $n_{j+1}/n_{j} \geq \rho > 1$ for all $j$ for some constant $\rho$, then $\lim_{r\uparrow 1} \sum_{j=0}^{\infty} a_j r^{n_j} = A$ if and only if $\sum_{j=0}^{\infty} a_j = A$. In particular, since $\sum_{n=0}^{\infty} (-1)^n$ does not converge, neither is $\lim_{r\uparrow 1} F(r)$. – Sangchul Lee Nov 22 '11 at 13:18

Note that $$ F(r)=r-F(r^2)\tag{1} $$ Thus, if $a=\lim\limits_{r\to1^-}F(r)$ exists, then $$ a=\lim_{r\to1^-}F(r)=\lim_{r\to1^-}r-\lim_{r\to1^-}F(r^2)=1-a\tag{2} $$ Therefore, if the limit exists then it is $a=\frac{1}{2}$.

Applying equation $(1)$ twice, we get $$ F(r)=r-r^2+F(r^4)\tag{3} $$ As $r\to1$, $(3)$ indicates $F$ tends toward being periodic in $-\log(-\log(r))$ with period $\log(4)$. Note that as $r\to1^-$, $-\log(-\log(r))\to\infty$. $F(r)$ is the sum of the lengths of the intervals in the following animation

enter image description here

The value of the sum oscillates between $0.49728$ and $0.50272$ over each period. Therefore, $\lim\limits_{r\to1^-}F(r)$ does not exist.

share|cite|improve this answer
What does it mean for a function to tend toward being periodic? – Did Nov 24 '11 at 15:48
@Didier: Let $t=-\log(\log(r))$ and define $G(t) = F(\exp(-\exp(-t)))$; i.e. $G(t)=F(r)$. Then, equation $(3)$ becomes $$G(t)=\left(e^{-e^{-t}}-e^{-e^{-t+\log(2)}}\right)+G(t-\log(4))$$ and as $t\to\infty$, $e^{-e^{-t}}-e^{-e^{-t+\log(2)}}\to0$. In other words, $$\lim_{t\to\infty}G(t)-G(t-\log(4))=0$$ That is the sort of almost periodicity I was meaning. – robjohn Nov 24 '11 at 17:19
In this sense, every function with a finite limit at infinity tends toward being periodic... – Did Nov 24 '11 at 21:24
Yes, every function with a finite limit at infinity would tend toward being a constant function (and constant functions are indeed periodic). However, my statement is attempting to describe how the limit fails to exist. $F$ oscillates between $0.49728$ and $0.50272$ infinitely often as $r\to1^-$. – robjohn Nov 25 '11 at 9:46

My question was connected with this one. Namely, consider the sequence $(1, -1, -1, 1, 1, 1, 1, -1, \dots)$, where $(-1)^{k}$ stands for indices from $2^{k-1}$ to $2^k-1$. The Cesaro means can easily be calculated and they don't have a limit. The function $F$ here corresponds to the Abel means, and the equivalence of these summation methods for the bounded sequence implies that the Abel means diverge, too.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.