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This question already has an answer here:

for two positive numbers $a_1 < b_1$ define recursively the sequence $a_{n+1} = \sqrt{a_nb_n}, b_{n+1} = \frac{a_n + b_n}{2}$. Show that $a_n, b_n$ converge to a common limit. Hint use inequality: $\sqrt{ab} \leq \frac{a+b}{2}$

attempt.

Suppose $a_n\longrightarrow L_1$ and $b_n \longrightarrow L_2$

using the hint:

$$\lim_{n \to \infty} 4a_nb_n < \lim_{n \to \infty} (a_n + b_n)^2$$

by properties of limits:

$$4(\lim_{n \to \infty} a_n)(\lim_{n \to \infty} b_n) <(\lim_{n \to \infty}a_n +\lim_{n \to \infty}b_n)^2$$

$$4L_1L_2 \leq (L_1 + L_2)^2$$

$$0 \leq (L_1 - L_2)$$ after taking the square root from each side. Then this implies

$$L_2 \leq L_1$$

Since there is an equality in the less than equal to expression then the limits can be equal and as such they converge to a common limit.

I didn't have any ideas of what else to try hopefully I was not reckless with my attempt.

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marked as duplicate by Arnaud D., Lord Shark the Unknown, mrtaurho, José Carlos Santos calculus Mar 15 at 13:52

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  • $\begingroup$ You can't take the square root . It's like saying $(1-3)^2 \geq 0$ so $1 \geq 3$ . $\endgroup$ – user252450 Feb 15 '16 at 20:05
  • $\begingroup$ Well......so much for that idea.......sigh $\endgroup$ – dc3rd Feb 15 '16 at 20:11
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The inequality $\sqrt{ab}\leq \frac{a+b}{2}$ implies that $a_n\leq b_n$ for all $n$. This in turn implies that $$ b_{n+1}=\frac{a_n+b_n}{2}\leq b_n $$ Therefore $\{b_n\}$ is a decreasing sequence of positive real numbers, hence has a limit b.

Next, $$ a_{n+1}=\sqrt{a_nb_n}\geq \sqrt{a_n^2}=a_n $$ Therefore $\{a_n\}$ is an increasing sequence. It is bounded above because $a_n\leq b_n$ for all $n$ and $\{b_n\}$ is bounded, so it has a limit $a$.

Finally, take $n\to\infty$ in both sides of $b_{n+1}=\frac{a_n+b_n}{2}$ to obtain $b=\frac{a+b}{2}$, or $a=b$.

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  • $\begingroup$ How does the original inequality imply $a_n \leq b_n$? If i manipulate it I don't get that $\endgroup$ – dc3rd Feb 15 '16 at 20:32
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    $\begingroup$ It comes from the recursive definition: $a_n=\sqrt{a_{n-1}b_{n-1}}\leq \frac{a_{n-1}+b_{n-1}}{2}=b_n$. $\endgroup$ – carmichael561 Feb 15 '16 at 21:07
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By the hint we get $a_n\leq b_n$ for every $n\in\mathbb{N}$, we also see that $a_n$ is monotonous increasing and $b_n$ is decreasing, so both sequences are monotonous and bounded $a_n\leq b_1$, $b_n\geq a_1$ and so they are convergent. Now we get by properties of limit $$\lim_{n\to\infty}b_n=\lim_{n\to\infty} b_{n+1}=\lim_{n\to\infty}\frac{a_n+b_n}{2}=\frac{1}{2}\left(\lim_{n\to\infty} a_n+\lim_{n\to\infty} b_n\right)$$ and so $$\frac{1}{2}\lim_{n\to\infty} a_n=\frac{1}{2}\lim_{n\to\infty} b_n $$ and the limits are the same.

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