Alternating lacunary series and 1-1+1-1+1-1+…

The series {1-1+1-1+1-1+\cdots} diverges. A common way to make it convergent is to replace each {1} with a power of {x}; the new series will converge when {|x|<1} and maybe its sum will have a limit as {x\rightarrow 1}. And indeed,

\displaystyle    x-x^2+x^3-x^4+x^5-x^6+\cdots = \frac{x}{1+x}

which tends to {1/2} as {x} approaches {1} from the left.

x/(1+x) has limit 1/2
x/(1+x) has limit 1/2

Things get more interesting if instead of consecutive integers as exponents, we use consecutive powers of {2}:

\displaystyle    f(x) = x-x^2+x^4-x^8+x^{16} -x^{32} +\cdots

On most of the interval {(0,1)} it behaves just like the previous one:

Lacunary series on (0,1)
Lacunary series on (0,1)

But there appears to be a little blip near {1}. Let’s zoom in:

Lacunary series near 1
… near 1

And zoom in more:

... and very near 1
… and very near 1

Still there.

This function was considered by Hardy in 1907 paper Theorems concerning infinite series. On pages 92–93 he shows that it “oscillates between finite limits of indetermination for {x=1}“. There is also a footnote: “The simple proof given above was shown to be by Mr. J. H. Maclagan-Wedderburn. I had originally obtained the result by means of a contour integral.”

Okay, but what are these “finite limits of indetermination”? The Alternating Series Estimation shows {0<f(x)<1} for {x\in (0,1)}, but the above plots suggest that {f} oscillates between much tighter bounds. Let’s call them {A= \liminf_{x\rightarrow 1-} f(x)} and {B=\limsup_{x\rightarrow 1-} f(x)}.

Since {f(x)+f(x^2)\equiv x^2}, it follows that {f(x)+f(x^2)\rightarrow 1} as {x\rightarrow 1^-}. Hence, {B = \limsup_{x\rightarrow 1-}(1-f(x^2)) = 1-A}. In other words, {A} and {B} are symmetric about {1/2}. But what are they?

I don’t have an answer, but here is a simple estimate. Let {g(x)=x-x^2} and observe that

\displaystyle    f(x) = g(x) + g(x^4)+g(x^{16}) + g(x^{64})+\dots \qquad \qquad (1)

The function {g} is not hard to understand: its graph is a parabola.

Yes, a parabola
Yes, a parabola

Since {g} is positive on {(0,1)}, any of the terms in the sum (1) gives a lower bound for {f}. Each individual term is useless for this purpose, since it vanishes at {1}. But we can pick {n} in {g(x^{4^n})} depending on {x}.

Let {x_0\in(0,1)} be the unique solution of the equation {x_0^4=1-x_0}. It could be written down explicitly, but this is not a pleasant experience; numerically {x_0\approx 0.7245}. For every {x>x_0} there is an integer {n\ge 1} such that {x^{4^n}\in [1-x_0,x_0]}, namely the smallest integer such that {x^{4^n} \le x_0}. Hence,

\displaystyle f(x)> g(x^{4^n})\ge \min_{[x_0,1-x_0]}g = x_0-x_0^2>0.1996 \qquad \qquad (2)

which gives a nontrivial lower bound {A>0.1996} and symmetrically {B<0.8004}. Frustratingly, this falls just short of neat {1/5} and {4/5}.

One can do better than (2) by using more terms of the series (1). For example, study the polynomial {g(t)+g(t^4)} and find a suitable interval {[t_0^4,t_0]} on which its minimum is large (such an interval will no longer be symmetric). Or use {3,4,5...} consecutive terms of the series… which quickly gets boring. This approach gives arbitrarily close approximations to {A} and {B}, but does not tell us what these values really are.

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