Proof the the Arithmetic-Harmonic Mean is expressible as the Geometric Mean

Question

We define the Arithmetic-Harmonic mean of $a,b \in \mathbb{R_+}$ such that \begin{gather*} a_{n+1} = \frac{1}{2}(a_n + b_n) \\ b_{n+1} = \frac{2a_{n}b_{n}}{a_{n} + b_{n}} \end{gather*}

Let us also assume that $a \neq b$. I am trying to prove that the limit of these sequences exist, they are equal to each other, and as $n \rightarrow \infty$, the limit is equal to $\sqrt{ab}$, the Geometric Mean of a,b.

This is stated by mathworld here: http://mathworld.wolfram.com/Arithmetic-HarmonicMean.html

Since I have proven that AM $\geq$ HM, with equality only holding at a = b, we can see that $a_1 \geq a_{2} \geq ... \geq a_n$. In the same vein we have $b_1 \leq b_{2} \leq ... \leq b_n$.

How can I rigourously show that these are monotonic sequences bounded above and below resepctively and therefore converge to each other?

Secondly, can anyone give me a hint as to how I can show that this sequence converges to the Geometric Mean?

Answer 1

Since $a_n b_n = a_{n+1} b_{n+1}$, in order to prove that the arithmetic-harmonic mean is the geometric mean we just need to prove that both the sequences $\{a_n\}_{n\geq 1},\{b_n\}_{n\geq 1}$ are converging to the same limit. By the AM-HM inequality:

$$ b_n \leq b_{n+1} \leq a_{n+1} \leq a_n \tag{1}$$ while: $$ a_{n+1}-b_{n+1} = \frac{a_n+b_n}{2}-\frac{2a_n b_n}{a_n+b_n} = \frac{(a_n-b_n)^2}{2(a_n+b_n)}\leq\frac{a_n-b_n}{2},\tag{2}$$ so that $\lim_{n\to +\infty}a_n = \lim_{n\to +\infty}b_n = \sqrt{a_1 b_1}$ as wanted.

Answer 2

Hint: The geometric mean possess the remarkable property that $G(x,y)=G(A,H)$, where A and H are the arithmetic and harmonic means, respectively, of x and y.

Answer 3

I will rewrite your question like this: "Let $a,b\in \mathbb{R}^+$, we define inductively sequences $x_n$ and $y_n$, $x_1=\sqrt{ab}$, $y_1=\frac{a+b}{2}$, $x_{n+1}=\frac{\sqrt{x_n y_n}}{2}$ and $y_{n+1}=\frac{x_n+y_n}{2}$. Then $x_n$ and $y_n$ are convergent and have the same limit."

We only need i simple lemma: Everey monotone bounded sequence is convergent. More details you can see : http://en.wikipedia.org/wiki/Monotone_convergence_theorem Suppose $a<b$, and by the AM_GM inequality $a<x_1<x_2<x_3<\ldots<y_3<y_2<y_1<b$. By the lemma above These sequence are convergent(hence have a limit). Now say $x=\lim x_n$ and $y=\lim y_n$. See that $y_{n+1}=\frac{x_n+y_n}{2}$ take the limit at both sides we get:

$y=\frac{x+y}{2}\Rightarrow y=x$.