# 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.

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:

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

And zoom in more:

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 me 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 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.

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.

## 2 thoughts on “Alternating lacunary series and 1-1+1-1+1-1+…”

1. One typo of no real importance: Your $\frac{1}{1+x}$ has the $x^5$ term occur twice at the start.

1. L says:

Thanks, fixed now.

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