Here is a cute problem from Peter Winkler.
The series below converges in the interval $[0,1)$
$$ x - x^2 + x^4 - x^8 + \dots + (-1)^n x^{2^n} + \dots $$
For $x \in [0, 1)$, call the value $f(x)$.
The question is, does the limit
$$ \lim_{x \to 1-} f(x) $$
exist?
Solution.
A proof by Noam Elkies goes as follows (rephrased, not verbatim):
We can show that for $x \in [0,1)$,
$$f(x) + f(x^2) = x $$
and so (by a repeated application of the above)
$$ f(x) = x - x^2 + f(x^4)$$
By the first identity, if the limit exists, it must be $\frac{1}{2}$.
Now for $0 \lt x \lt 1$ we must have that (using the second identity above)
$$ f(x) \gt f(x^4) \gt f(x^{16}) \gt \dots \gt f(x^{4^n}) \gt \dots$$
Thus if there was some $c$ such that $f(c) \gt \frac{1}{2}$, we must have that
$$f(c^{4^{-n}}) \gt f(c) \gt \frac{1}{2}$$
And since $c^{4^{-n}} \to 1$ as $n \to \infty$, the limit cannot be $\frac{1}{2}$ if $f(c) \gt \frac{1}{2}$.
A quick computation (wrote python code) upto $15$ terms shows that $f(0.9999) \gt 0.5006$.
