Different way to calculate a different mean

The arithmetic-harmonic mean converges to the geometric mean. This can be seen by the invariance of the GM after each iteration:

$\sqrt{a_{n}h_{n}} = \sqrt{\frac{a_{n}+h_{n}}{\frac{a_{n}+h_{n}}{a_{n}h_{n}}}} = \sqrt{\frac{a_{n}+h_{n}}{\frac{1}{a_{n}}+\frac{1}{h_{n}}}} = \sqrt{\frac{a_{n}+h_{n}}{2} \frac{2}{\frac{1}{a_{n}}+\frac{1}{h_{n}}}} = \sqrt{a_{n+1}h_{n+1}}$

So what we're dealing with is the nested radical

$\sqrt{2^{1}\sqrt{2^{4}\sqrt{2^{9}\sqrt{\ldots}}}} = 2^{1^{2}/2^{1}} \cdot 2^{2^{2}/2^{2}} \cdot 2^{3^{2}/2^{3}} \cdot \ldots$

Taking log base $2$ gives us the infinite sum

$\sum_{n=1}^{\infty} \frac{n^{2}}{2^{n}}$

It is well-known that $\displaystyle \sum_{n=1}^{\infty} x^{n} = \frac{x}{1-x}$ for all $|x| < 1$ . Differentiating both sides and multiplying by $x$ gives us the identity $\displaystyle \sum_{n=1}^{\infty} nx^{n} = \frac{x}{(1-x)^{2}}$ . Again differentiating and multiplying $x$ gives us $\displaystyle \sum_{n=1}^{\infty} n^{2}x^{n} = \frac{x(1+x)}{(1-x)^{3}}$ .

Since $x = \frac{1}{2}$ satisfies that $|x| < 1$ , we can use the above identity to evaluate our infinite series:

$\sum_{n=1}^{\infty} \frac{n^{2}}{2^{n}} = \frac{\frac{1}{2}(1+\frac{1}{2})}{(1-\frac{1}{2})^{3}} = 6$

Hence, the nested radical turns out to be $2^{6} = \boxed{64}$ . Hurrah!