Limit of sequence $a_1=1$ and $a_{n+1}=\frac{2a_n}{4a_n+1}$

Question

I'm reading Sudhir R. Ghorpade and Balmohan V. Limayes: A Course in Calculus and Real Analysis and ran into a small issue with exercise 9 iii of chapter 2.

We are asked to find the limit of the following sequence

$$a_1=1$$

$$a_{n+1}=\frac{2a_n}{4a_n+1}$$

To start, I show that

$$a_{n+1}-a_n = \frac{2(a_n-a_{n-1})}{(4a_n+1)^2}$$

Now, since $a_n \leq 1$, we have that $$4a_n+1 \leq 5$$

Which I then use to derive the inequality

$$|a_{n+1}-a_n| \leq \frac{2}{25}|a_n - a_{n-1}|$$

This, by an earlier result in the chapter, implies the sequence is Cauchy and therefore convergent.

Let $a_n \to a$, then $a_{n+1} \to a$ and we can rewrite $$a_{n+1}=\frac{2a_n}{4a_n+1}$$ $$a=\frac{2a}{4a+1}$$

Solving for a gives the solutions $0$ and $1/4$.

How do I know which one is the right one?

Answer 1

Note that if $a_n\geqslant\frac14$, then$$a_{n+1}-\frac14=\frac{2a_n}{4a_n+1}-\frac14=\frac{4a_n-1}{4(4a_n+1)}\geqslant0.$$So, $a_{n+1}\geqslant\frac14$. Since $a_1>\frac14$, this proves that $(\forall n\in\mathbb{N}):a_n\geqslant\frac14$. Therefore, the limit is $\frac14$.

Answer 2

Let $b_n=\frac1{a_n}$ (possible because the recursion formula cannot produce zero from non-zero input). Then $$b_{n+1}=\frac{4a_n+1}{2a_n} =2+\frac1{2a_n}=2+\frac12b_n.$$ I suppose it is easier to see that $b_n\to 4$ (or at least that $b_n$ is bounded and hence $a_n\not\to 0$). Indeed, $$ b_{n+1}-4=-a+\frac12b_n=\frac12(b_n-4)$$ so that $$b_n-4=2^{-n}(b_0-4). $$