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We have the following for $a \le b \le c >0$:

$A(a,b,c)=\frac{a+b+c}{3}, B(a,b,c)= (abc)^{1/3}, C(a,b,c)=\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}} $.

Then we define the sequences $(a_n),(b_n), (c_n)$ by

$a_1=a, b_1=b, c_1=c,$

$a_{n+1}=A(a_n,b_n,c_n), b_{n+1}=B(a_n,b_n,c_n), c_{n+1}=C(a_n,b_n,c_n)$.

How can we show that $(a_n),(b_n), (c_n)$ are convergent and have the same limit?

I understand that $A(a,b,c)\ge B(a,b,c)$ and $B(a,b,c)\ge C(a,b,c)$

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  • $\begingroup$ @dxiv I think so. $\endgroup$ – MOP Feb 6 '18 at 5:13
  • $\begingroup$ See this and this related questions, same idea applies here. $\endgroup$ – dxiv Feb 6 '18 at 5:17
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    $\begingroup$ Assume you meant $0 < a \le b \le c$, and not that only $c$ is positive. $\endgroup$ – Macavity Feb 6 '18 at 6:27
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We have $$ A(x,y,z) = \mbox{ arithmetic mean of } x,y,z; $$ $$ B(x,y,z) = \mbox{ geometric mean of } x,y,z; $$ $$ C(x,y,z) = \mbox{ harmonic mean of } x,y,z. $$

It is well known that, for the same arguments, $$ \mbox{ harmonic mean } \le \mbox{ geometric mean } \le \mbox{ arithmetic mean } \tag{1} $$ (see e.g. this Wikipedia article).

Using these inequalities we see that the largest (arithmetic) means $a_n$ form a non-increasing sequence, while the smallest (harmonic) means $c_n$ form a non-decreasing sequence. Both sequences are bounded: all terms are within the interval $[a,c]$. If a sequence is monotonic and bounded, it has a limit.

Can you finish by proving that the limits of $a_n$ and $c_n$ are the same? (Then $b_n$ necessarily has the same limit too because of the double inequality $(1)$ and the squeeze theorem.)

For $n>1$ we have $$ a_{n+1}={a_n+b_n+c_n\over3}\le{a_n+a_n+c_n\over3}, \tag{2} $$ $$ c_{n+1} \ge c_n. \tag{3} $$ Subtracting $(3)$ from $(2)$ we find $$ a_{n+1}-c_{n+1} \le {2\over3} (a_n-c_n). $$ We observe that the intervals $[c_n,a_n]$ form a sequence of nested closed intervals, and the interval lengths tend to zero (no slower than a decreasing geometric progression). Therefore these intervals have a (unique) common point, and this point must be the limit of all three sequences because $a_n,b_n,c_n \in [c_n,a_n]$ for all $n>1$. (The common point is unique because the size of intervals $[c_n,a_n]$ tends to zero.) This completes the proof.

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  • $\begingroup$ I can understand that $(a_n)$ is decreasing sequence and $(c_n)$ is a increasing sequence. But could you please please help me with finding the limit? $\endgroup$ – MOP Feb 6 '18 at 5:32
  • $\begingroup$ You are not required to find the limit. You just need to prove that the limits are the same. $\endgroup$ – Alex Feb 6 '18 at 5:37
  • $\begingroup$ For example, you can prove that $a_{n+1}-c_{n+1}\le0.9(a_n-c_n)$, and then observe that the limit of each sequence is in the interval $[c_n,a_n]$ for all $n\in{\mathbb N}$. Note that the intervals $[c_n,a_n]$ form a sequence of nested intervals. $\endgroup$ – Alex Feb 6 '18 at 5:53
  • $\begingroup$ I could't find $a_{n+1}-c_{n+1} \le \frac{1}{x} (a_n-c_n)$. I can understand it for two numbers case only. $\endgroup$ – MOP Feb 6 '18 at 7:48
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hint: try proving $c_n$ is increasing and $a_n$ is decreasing and use squeeze theorem.

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Assuming $a>0.$ For any positive $a',b',c'$ we have $$\max (a',b',c')\geq A(a',b',c')\geq B(a',b','c)\geq C(a',b',c')\geq \min (a',b',c').$$ Let $U_n=\max (a_n,b_n,c_n)$ and $L_n=\min (a_n,b_n,c_n).$ $$\text {We have }\quad U_n\geq U_{n+1}\geq L_{n+1}\geq L_n.$$ So $(U_n)_n$ is a deceasing sequence bounded below by $L_1$ and $(L_n)_n$ is an increasing sequence bounded above by $U_1 .$

Let $U=\lim_{n\to \infty}U_n$ and $L=\lim_{n\to \infty}L_n.$ Obviously $U\geq L\geq L_1>0.$

It suffices to show that $U=L.$

Let $r_n=U_n-L_n .$ Note that $U_n>r_n.$ $$\text {We have }\quad U_{n+1}\leq \frac {2U_n+L_n}{3}$$
$$\text {and }\quad L_{n+1}\geq \frac {3}{\frac {1}{U_n}+\frac {2}{L_n}}=\frac {3U_nL_n}{2U_n+L_n}.$$ $$\text {Therefore }\quad r_{n+1}\leq \frac {2U_n+L_n}{3}-\frac {3U_nL_n}{2U_n+L_n}=$$ $$=\frac {(4U_n-L_n)(U_n-L_n)}{3(2U_n+L_n)}=$$ $$=\frac {(3U_n+r_n)r_n}{9U_n-3r_n}\leq$$ $$\leq \frac {(4U_n)r_n}{6U_n}=\frac {2}{3}r_n.$$

So $U-L=\lim_{n\to \infty}r_n=0$ because $0\leq r_{n+1}\leq \frac {2}{3}r_n.$

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  • $\begingroup$ How did you show that $a_n$ is increasing and $b_n$ is increasing? $\endgroup$ – MOP Feb 6 '18 at 10:33
  • $\begingroup$ I didn't understand your proof? Why did you take U and L? $\endgroup$ – MOP Feb 6 '18 at 11:22
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    $\begingroup$ $U_n=\max (a_n,b_n,c_n).$... $L_n=\min (a_n,b_n,c_n).$... $U_{n+1}= \max (a_{n+1},b_{n+1},c_{n+1}) =$ $ \max (\; A(a_n,b_n,c_n), B(a_n,b_n,c_n),C(a_n,b_n,c_n)\;)\;=$ $A(a_n,b_n,c_n)\leq \max (a_n,b_n,c_n)=U_n.$..... For $n>1$ we have $ U_n=a_n$ and $L_n=c_n.$.... So $a_n$ is decreasing, not increasing..... If $U_n=\max (a_n,b_n,c_n)$ converges to $U$ and if $L_n= \min (a_n,b_n,c_n)$ converges to $L$ and if $U=L$ then $a_n, b_n,c_n$ each converge to $L$ also. $\endgroup$ – DanielWainfleet Feb 6 '18 at 23:18

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