Finding a limit of a monotonically decreasing function Let $a_n$ be a series so $a_1=1$ and $2a_{n+1}<a_n<3a_{n+1}$. Find the limit of $a_n$ when $n\to \infty$.
I have proved that $a_n>0$ and that it monotonically decreasing. But how I find the limit now? Should I use the Squeeze theorem? I know how to solve it when for example we have $a_n = -a_{n+1} + 1$ (we do: $L=\lim_{n\to \infty}a_n $ and $L=-L+1$). But I don't have an equation here.
 A: Suppose that the limit exists and we call it $L$. We then have
$$\lim_{n\to\infty}2a_{n+1}\leq\lim_{n\to\infty}a_n\implies 2L\leq L$$
In other words, $L\leq 0$. Similarly, using the upper bound, we can say $L\leq 3L$, so $L\geq 0$. Thus, if the limit exists then it must be $0$. Since you have shown that the sequence is monotonically decreasing and bounded below, then it must converge, so it must converge to $0$.
A: $$2L \leq L \leq 3L$$
If $L>0$ then $2L \leq L$ is absurd. If $L<0$ then $L \leq 3L$ is absurd, so $L=0$.
A: Since $a_n>0$ for all $n$ and since the sequence is monotonically decreasing, then by Monotone Convergence Theorem, there is a real number $L$ such that $\lim_{n\to\infty}a_n=L.$
By standard results, we then have that $$\lim_{n\to\infty}2a_{n+1}=2L$$ and $$\lim_{n\to\infty}3a_{n+1}=3L,$$ and moreover that $$2L\le L\le 3L.$$
Thus, $$0\le-L\le L.$$ Since neither $L$ nor its opposite is negative, then $L=0.$

Alternatively, I suspect that you've already noted that $$a_{n+1}<\frac12 a_n$$ for all $n.$ An inductive proof readily shows that $$\frac1{2^n}a_1=\frac1{2^n}$$ for all $n.$ Thus, $0<a_{n+1}<\frac{1}{2^n}$ for all $n,$ and so the Squeeze Theorem lets us conclude that $$0\le\lim_{n\to\infty}a_{n+1}\le\lim_{n\to\infty}\frac{1}{2^n}=0,$$ at which point, we're done, since $\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}a_n.$
