<s>[INST] Elementary lemma about a rational sequence This was an exercise to prove the Lemma below. I think there may be some easy theorem that does this more quickly. Any corrections or suggestions appreciated.

Lemma. Let $\{a_n\}$ be a sequence of non-negative real numbers, an infinite number of which are non-zero. If $\lim_{n \rightarrow \infty} \frac{a_n}{a_n+a_{n+1} } = L$ exists, then either
(a) $L = 1/2,$ (b) $ a_n$ grows exponentially, or (c) $\frac{1}{a_n }$ grows exponentially, and in any case only finitely many of the $a_n$ are zero.

Proof. $0 \leq \frac{a_n}{a_n+a_{n+1}}\leq 1, $ so $0\leq L \leq 1.$
A. First, if $L = 0, a_n$ grows exponentially. Since $ L = 0, \exists N $ such that for all $n \geq N$ either  $$  \frac{a_n}{a_n + a_{n+1}}< 1/3$$ or $$a_n + a_{n+1 }= 0.$$ The latter can happen only finitely many times since we assume the limit exists. So $$ a_n \leq \frac{1}{3}(a_n+a_{n+1}),~~ \forall n\geq N.$$  This gives $\frac{2}{3}a_n \leq \frac{1}{3}a_{n+1}$ and so $a_{n+1} \geq 2a_n $ and then finally $ \forall m > 0, \forall n \geq N,$ $$a_{n+m} \geq 2^m a_n.$$
Fixing n such that $a_n \neq 0$ gives that $\{ a_n\}$ grows exponentially.
This also shows that $a_{n+m} > 0,~~ \forall m > 0,$ so only a finite number of $a_n$ can be zero.
B. Now assume $L\neq 0.$
First, there are only finitely many $a_n$ with $a_n = 0.$ Otherwise we would have infinitely many $\frac{a_n}{a_n + a_{n+1}} = 0, $ which contradicts our assumption that $L\neq 0.$
Since there are only finitely many $a_n = 0$ we need not be concerned about zero denominators $a_n + a_{n+1}= 0. $
Now assume $L< 1/2 $ and let $\epsilon = \frac{1}{4} - \frac{L}{2} > 0. $ Then $ \frac{a_n}{a_n+ a_{n+1}}< L + \epsilon.$
$$a_n < (L+\epsilon )(a_n + a_{n+1}) \implies (1 - L - \epsilon) a_n < (L + \epsilon)a_{n+1}$$ $$\implies (\frac{1}{L+\epsilon}- 1)a_n < a_{n+1} \implies (\frac{1}{L+ 1/4 - L/2)}-1)a_n< a_{n+1} $$ $$\implies (\frac{1}{L/2 + 1/4} -1) a_n < a_{n+1} \implies (\frac{1}{L/2+ 1/4} -1)^m a_n < a_{n+m}.$$ Since $L < 1/2, (\frac{1}{L/2 + 1/4}-1) > 1.$
Therefore $a_n $ grows exponentially.
If $L > 1/2,$ then we consider the sequence $b_n = 1/a_n.$
We have $$\lim_{n \rightarrow \infty} \frac{b_n}{b_n + b_{n+1}} = \lim \frac{1/a_n}{1/a_n + 1/a_{n+1}} = \lim \frac{a_{n+1}}{a_n + a_{n+1}} = 1 - \lim_{n \rightarrow \infty} \frac{a_n}{a_n+a_{n+1}} = 1 - L < 1/2.$$
By the argument above $b_n$ too should grow exponentially.
 [/INST] $\frac {a_{n+1}} {a_n}$ can be written as a function of $\frac{a_n}{a_n + a_{n+1}}$ :
$\frac {a_n}{a_n + a_{n+1}} =\frac 1 {1 + a_{n+1}/a_n}$. Inverting $x \mapsto 1/(1+x)$ gives you $\frac {a_{n+1}}{a_n} = \frac {1 - \frac{a_n}{a_n + a_{n+1}}}{\frac{a_n}{a_n + a_{n+1}}}$
Now stating the lemma in terms of $\frac {a_{n+1}}{a_n}$ instead of $\frac{a_n}{a_n + a_{n+1}}$, it becomes :
If $\lim \frac {a_{n+1}}{a_n} = L$ exists, then either (a) $L = \frac {1-1/2}{1/2} = 1$ (b) $a_n$ grows exponentially (in fact when $L > 1$), or (c) $1/a_n$ grows exponentially (when $0 < L < 1$).
Which should be easy to prove.
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