TITLE: Limit of sequence $a_1=1$ and $a_{n+1}=\frac{2a_n}{4a_n+1}$
QUESTION [7 upvotes]: 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?

REPLY [5 votes]: 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 $\frac14$.

REPLY [3 votes]: 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 $$b_n-4=2^{-n}(b_0-4). $$