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.


Use the squeeze principle.

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    $\begingroup$ A somewhat different hint that simplifies the problem a bit: define sequences $c_n$ and $d_n$ by $c_n=\frac1{a_n}$ and $d_n=\frac1{b_n}$; you should be able to convert the expression for $a_n$ into a much simpler expression for $c_n$. A magic phrase is 'arithmetic-geometric mean'. $\endgroup$ – Steven Stadnicki Aug 20 '13 at 5:02
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    $\begingroup$ And after that transformation, see this question. $\endgroup$ – Calvin Lin Aug 20 '13 at 5:26
  • $\begingroup$ See also Wikipedia: Arithmetic-geometric mean. (Limit from this question could be described as "arithmetic-harmonic mean".) $\endgroup$ – Martin Sleziak Aug 20 '13 at 8:05

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.


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