# 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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