| It was a bit trickier to find a simple solution here than for the case of sums.
First, note that for x=0 any series with mu(0)=0 will do, and if we can solve
the problem for x>0, we can change the sign of mu(0) to get a solution for -x.
So WLOG x>0. We will always choose nonzero mu(i) with the same sign as a(i); so
WLOG we may also assume all a(i)>0 as well. Now we can work more easily since
all relevant quantities will be positive.
Next, since a(n)->0, I wish to choose a point k beyond which all a(i)<1/2, and
let a(i)=1 before this point; this reduces the problem to a new one in which x'
is x divided by the product of the initial a(i), and a'(i)=a(i-k).
Let P(n) be product(i in 1..n)mu(i)*a(i), so P(n)=mu(n)*a(n)*P(n-1) for n>0, and
P(-1)=1 by convention. Define inductively mu(n)=ceiling(x/(a(n)*P(n-1)). (The
value of mu(n) depends on previous mu(i), but this is well-defined.) It is
clear that mu(n) is always a positive integer.
We have mu(n) >= x/(a(n)*P(n-1)); multiplying by a(n)*P(n-1) and using the
definition of P(n) gives us P(n)>=x for all n>=0. Similarly,
mu(n) < 1 + x/(a(n)*P(n-1)), so
P(n) < a(n)*P(n-1) + x = P(n)/mu(n) + x.
Using the first part, P(n) < a(n)*P(n-1) + x < P(n-1)/2 + x, using a(n)<1/2; by
induction this gives us P(n)<2(x+1) for n>=-1, so the P(n) are bounded above.
Using the second part, P(n) < x/(1-1/mu(n)) when mu(n)/=1. From the definition
of mu(n) and from having the P(n) bounded and a(n)->0, we get mu(n)->infinity.
So 1-1/mu(n)->1 and P(n) is sandwiched between x and a value that approaches x.
So P(n)->x.
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