4
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.