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Define our Harmonic sequence for two numbers such that \begin{equation} a_{n+1} = \frac{2a_nb_n}{a_n + b_n} \end{equation} and our geometric sequence \begin{equation}b_{n+1} = \sqrt{a_nb_n} \end{equation} such that as $n \rightarrow \infty$ we tend towards the Geometric-Harmonic Mean.

The arithmetic-geometric mean can be defined by the following two sequences. First compute the arithmetic mean of two numbers $a,b \in \mathbb{R_+}$ such that $a_1$ = $\frac{1}{2}(a + b)$ and then compute the geometric mean such that $b_1$ = $\sqrt{ab}$. If we continue to iterate this operation we can define our two sequences {$a_n$} and {$b_n$} as follows: \begin{gather*} a_{n+1} = \frac{1}{2}(a_n + b_n) \\b_{n+1} = \sqrt{{(a_nb_n)}}. %Try and get this all under 1 square root sign \end{gather*} As ${n\to\infty}$ we approach our arithmetic-geometric mean (agM).

I have proved that the Geometric-Harmonic Mean exists and that $a_n$ = $b_n$ as $n \rightarrow \infty$. However, Mathworld states without proof that the limit for our Geometric-Harmonic mean can be written as \begin{equation*} \lim_{n \to \infty}a_n = \frac{1}{M({a_n}^{-1},{b_n}^{-1})} (*) \end{equation*}

Where M is the arithmetic geometric mean. You can find the link here: http://mathworld.wolfram.com/Harmonic-GeometricMean.html

Can anyone help me prove (*)?

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Let us suppose that as per your definition $\{a_{n}\}, \{b_{n}\}$ form Harmonic-geometric sequences. Let $A_{n} = 1/a_{n}, B_{n} = 1/b_{n}$ Then clearly we have $$B_{n + 1} = \sqrt{A_{n}B_{n}}$$ and \begin{align} A_{n + 1} &= \frac{1}{a_{n + 1}}\notag\\ &= \frac{a_{n} + b_{n}}{2a_{n}b_{n}}\\ &= \dfrac{\dfrac{1}{A_{n}} + \dfrac{1}{B_{n}}}{\dfrac{2}{A_{n}B_{n}}}\notag\\ &= \frac{A_{n} + B_{n}}{2}\notag \end{align} It follows that $\{A_{n}\}, \{B_{n}\}$ form Arithmetic Geometric sequence and hence both tend to the common limit which we denote by $M(A_{1}, B_{1}) = M(a_{1}^{-1}, b_{1}^{-1})$. Now $A_{n} = 1/a_{n}$ so that $a_{n} \to 1/M(a_{1}^{-1}, b_{1}^{-1})$. Thus the desired relation is proved.

It thus goes on to show that the harmonic geometric sequences don't lead to a new concept and behaves as reciprocal of arithmetic geometric sequences.

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  • $\begingroup$ Thank you! This solves my issue, and I would vote the answer up if I had but 15 reputation! $\endgroup$
    – rjadler
    Mar 15 '15 at 22:06

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