Confused how the substitution expresses this recurrence

Question

Consider the recurrence $a_n = a_{n-1}(1 - a_{n-1})$ for $n > 0$ with $a_0 = \frac{1}{2}$. Since the terms in the recurrence decrease and are positive, it is not hard to see that $\lim_{n\to\infty}a_n = 0$. To find the speed of convergence, it is natural to consider $1/a_n$. Substituting, we have $\frac{1}{a_n} = \frac{1}{a_{n-1}}(\frac{1}{1-a_{n-1}}) = \frac{1}{a_{n-1}}(1+a_{n-1}+a_{n-1}^2+\cdots) > \frac{1}{a_{n-1}} + 1$.

I don't understand how to get $ \frac{1}{a_{n-1}}(\frac{1}{1-a_{n-1}}) = \frac{1}{a_{n-1}}(1+a_{n-1}+a_{n-1}^2+\cdots)$. I think it was by factoring but can't seem simplify to that expression. Would anyone please show me how it happened?

The power series $\frac 1{1-x} = 1 + x + x^2 + x^3 + \dots$ converges when $|x|<1$. You may recognise it as the sum of an infinite geometric series. — player3236, Oct 06 '20 at 19:47
Why not use $$\frac1{a_n}=\frac1{a_{n-1}}+\frac1{1-a_{n-1}}>\frac1{a_{n-1}}+1?$$ — , Oct 06 '20 at 19:58
@ProfessorVector How does $\frac{1}{a_{n-1}} + \frac{1}{1-a_{n-1}} = \frac{1}{a_n}$? — dalton atwood, Oct 06 '20 at 22:54
@randomvalue Add the two fractions by using the common denominator $a_{n-1}(1-a_{n-1})=a_n$. — RobPratt, Oct 07 '20 at 02:22
Came from your other question. This statement is in the first sentence of your question. — player3236, Oct 07 '20 at 15:46

score 1 · Answer 1 · answered Oct 07 '20 at 01:55

That is just the geometric series. As you state, $a_n$ decreases (and starts out at $1/2$), so the expansion (valid for $\lvert a_n \rvert < 1$) works:

$\begin{align*} \frac{1}{1 - a_n} &= 1 + a_n + a_n^2 + \dotsb \\ &= \sum_{k \ge 0} a_n^k \end{align*}$

Confused how the substitution expresses this recurrence

1 Answers1