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?
Best Answer
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.