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Solution to puzzle 141: Alternating series

Skip restatement of puzzle.Consider the alternating series f(x) = x − x2 + x4 − x8 + ... + (−1)n x(2n) + ... , which converges for |x| < 1.  Does the limit of f(x) as x approaches 1 from below exist, and if so what is it?


Firstly, note that f(x) = x − f(x2).
If f(x) has limit L as x approaches 1 from below, then L = 1 − L, so that L = ½.
(That is, if the limit exists, then for any e > 0, there exists X such that for all x > X (with |x| < 1), |f(x) − ½| < e.)
We will show that the limit cannot be equal to ½, and hence the limit does not exist.

From here there are a number of routes we could take.  We might try to prove that the limit exists, perhaps using one of the standard tests for convergence.  Any such attempt would fail.  Or we might take a more computational approach, by calculating f(x) for various values of x close to 1.  By itself, this approach will not yield a proof, but it may provide some useful hints.

Note that, by the Comparison Test, f(x) is absolutely convergent for 0 less than or equal to x < 1: consider
g(x) = x + x2 + x4 + x8 + ... + x(2n) + ... , and
h(x) = x + x2 + x3 + x4 + ... + xn + ... ,
where h(x), a geometric series, is known to converge. 

Hence we may bracket the terms of f(x), and write f(x) = (x − x2) + (x4 − x8) + (x16 − x32) + ...
Note that all of the bracketed terms are positive, as 0 < a < b implies xa > xb, for 0 < x < 1.
This allows us to easily calculate a lower bound for f(x), for a given value of x.
The table below shows the sum of 16 terms (or eight bracketed terms) of f(x) for x = 0, 1/2, 3/4, 7/8, ... , 4095/4096, correct to six decimal places.  Notice that f(4095/4096) > ½.  We can be sure of this because we are taking the sum of a finite number of positive terms.

Lower bound for f(x) for selected values of x
mx = 1 − (½)mf(x): sum of 16 terms
000
11/20.308609
23/40.413715
37/80.456269
415/160.479453
531/320.489163
663/640.495031
7127/1280.497185
8255/2560.498879
9511/5120.499179
101023/10240.499838
112047/20480.499676
124095/40960.500078

Next, note that f(x) = x − x2 + f(x4).
Since x > x2 for 0 < x < 1, we have f(x) > f(x4).

Putting these results together, and setting a = 4095/4096, we find that:
½ < f(a) < f(a1/4) < f(a1/16) < f(a1/64) < ... .
The sequence {a, a1/4, a1/16, a1/64, ... } is strictly increasing, and approaches 1.

This shows explicitly that, if we take e = 0.00007, there is no value of X such that for all x > X, (with |x| < 1), |f(x) − ½| < e.  (No matter what value of X we choose, we can always find an element p > X of the sequence {a, a1/4, a1/16, a1/64, ... }, for which f(p) − ½ > e.)
Hence f(x) does not approach ½, and thus has no limit at all. 

Therefore, the limit of f(x) as x approaches 1 from below does not exist.


Remarks

It may be asked what f(x) does as x approaches 1.  The answer is that it oscillates infinitely many times, with each oscillation about four times quicker than the previous one.  See the solution given in the reference below for more details.

Source: Puzzle 8 on Noam's Mathematical Miscellany

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