# Different way to calculate a different mean

Calculus Level 5

You may have heard of the arithmetic-geometric mean $$\text{agm}(x,y)$$. Let us define the arithmetic-harmonic mean $$\text{ahm}(x,y)$$ as such:

Let $$x = a_{0}$$ and $$y = h_{0}$$, and define $$a_{n+1} = \frac{a_{n}+h_{n}}{2}$$ and $$h_{n+1} = \frac{2}{\frac{1}{a_{n}}+\frac{1}{h_{n}}}$$. Then $$\displaystyle \text{ahm}(x,y) = \lim_{n \rightarrow \infty} a_{n} = \lim_{n \rightarrow \infty} h_{n}$$.

Evaluate

$\text{ahm}(2^{1}, \text{ahm}(2^{4}, \text{ahm}(2^{9}, \text{ahm}(2^{16}, \ldots))))$

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