Let $a_1$ and $b_1$ be any two positive numbers, and define $\{ a_n\}$ and $\{ b_n\}$ by
$$a_n = \frac{2a_{n-1}b_{n-1}}{a_{n-1}+b_{n-1}},$$
$$b_n = \sqrt{a_{n-1}b_{n-1} }.$$
Prove that the sequences $\{a_n\}$ and $\{b_n\}$ converge and have the same limit.
Source: Problem Solving Through Problems by Loren C. Larson.
Hint:
Use the squeeze principle.
Best Answer
Let $A_n$ denote the arithmetic mean of $a_n$ and $b_n$, and $G_n$ their geometric mean. We have
$$a_{n+1} = \frac{G_n^2}{A_n} \leq G_n = b_{n+1}$$
If, WLOG, $a < b$, then
$$a_n < a_{n+1} < b_{n+1} < b_n$$
on inspection, which establishes the monotonicity and boundedness of both sequences; thus, they converge. In particular, since $a_n$ is Cauchy, we find (fixing $\epsilon$) that there is $N$ such that
$$\left| a_n - a_{n-1} \right| = \left| \frac{a_{n-1} (a_{n-1} - b_{n-1})}{a_{n-1} + b_{n-1}} \right| < \epsilon$$
for any $n > N$. Since $\frac{a_{n-1}}{a_{n-1} + b_{n-1}} < 1$, we see that
$$\left| (a_{n-1} - b_{n-1}) \right| < \epsilon$$ if $n > N$, which proves our claim.