I've come across this infinite series: $\sum_{n=0}^\infty x^{n^\alpha}$, with $0<x<1$ and $\alpha > 0$.
Does this series have a name and/or is there a method for computing it (besides brute force, obviously)? Thanks!
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2 Answers
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For $\alpha > 1$, this sum will converge extremely rapidly, even if $x$ is fairly close to $1$ (you can't express in floating point numbers those numbers close enough to $1$ where this would converge slowly). The only case that is difficult, convergence wise, is when $\alpha$ is very close to 0.
In that case, your best bet are Levin's U-transform and related algorithms. Even better, just link in to GSL, which already implements that.
answered Mar 17, 2010 at 17:00
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$\begingroup$ What do you mean by "For $n\geq2$"? $\endgroup$ Commented Mar 17, 2010 at 18:17
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$\begingroup$ And note the reference to the extensive review paper arxiv.org/abs/math/0005209 found in the second link above. $\endgroup$ Commented Mar 17, 2010 at 18:19
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$\begingroup$ @Jonas: I bet that's just a slip o' ye olde keyboard and shouldn't have been there. The answer seems to make good sense if you just delete it. $\endgroup$ Commented Mar 17, 2010 at 18:24
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$\begingroup$ Yep, that $n\geq 2$ was a simple mistake [I had too many variants of the equation in my Maple session where I 'played around' with it first before posting my answer. Error edited away, thanks] $\endgroup$ Commented Mar 17, 2010 at 18:53
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$\begingroup$ I thought so. I wouldn't mind all comments referring to it being deleted for posterity, but if I only delete mine it may be confusing. $\endgroup$ Commented Mar 17, 2010 at 19:12
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The $k$th power is the generating function for the ways of expressing $n$ as a sum of $k$ $\alpha$th powers of positive integers. However, I don't think this helps much even for most integer values of $\alpha$.
When $\alpha = 1$, this is a geometric series.
When $\alpha = 2$, this is a theta function related to the Jacobi triple product formula
$$\prod_{m=1}^\infty (1-x^{2m})(1+x^{2m-1}y^2)(1+x^{2m-1}y^{-2}) = \sum_{n=-\infty}^\infty x^{n^2}y^{2n}$$ since $$ (\frac 12 \text{RHS}+\frac12) \bigg|\_{y=1} = \sum\_{n=0}^\infty x^{n^2} .$$
You may be able to compute the sum when $\alpha=2$ more efficiently using one of the integral formulas or other properties for elliptic theta functions. Other than that, I don't know of special cases.
This doesn't say anything about whether some rapid series acceleration technique exists.
answered Mar 17, 2010 at 15:29
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1$\begingroup$ For $\alpha=2$ you can use the functional equation for the theta function to get a massive speed-up. Set $\theta(t)=\sum_ne^{-\pi n^2 t^2}$, where the sum is over all integers. This converges for $t>0$ real. Much less clear, but completely standard, is that $\theta(1/t)=t\theta(t)$. Using this trick you can reduce to the case $0<t\leq 1$ which corresponds to $0<x<0.0432...$ where the series is converging super-fast. $\endgroup$ Commented Mar 17, 2010 at 19:47
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$\begingroup$ As an explicit example, if $x=0.9$ and $\alpha=2$ then summing 1000 terms gives an answer of 3.230272... . But solving $e^{-t^2}=x$ gives t=0.183... so $u:=1/t=5.4605...$ and summing just one term for the series with $x=e^{-\Pi u^2}$ gives 1, correct to 40 or so dec places, so $\theta(t)$ is approximately $u$ (to about 40 dec places!) so the sum we're interested in is hence (1+u)/2. In summary, the answer with $x=0.9$ and $\alpha=2$ is very very close to $(1+u)/2$ with $u=(-\log(0.9)/\pi)^{-1/2}$. As $x$ gets closer to 1 this formula gets better and better. $\endgroup$ Commented Mar 17, 2010 at 19:55
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$\begingroup$ [...and if you want more accuracy then use more than one term when expanding $\theta(u)$!] $\endgroup$ Commented Mar 17, 2010 at 19:57
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$\begingroup$ [aargh slip above "reduce to the case $t\geq1$". When can we edit comments?? $\endgroup$ Commented Mar 17, 2010 at 19:59
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$\begingroup$ For $\theta(1/t)=t\theta(t)$ read $\theta(1/t)=\sqrt{t}\theta(t)$. (Not that this helps much with the original question). So again, when can we edit comments? $\endgroup$ Commented Mar 17, 2010 at 21:00