IMC 1995 · Day 2 · P8

Put $a_n = 1 + \sqrt{b_n}$ for $n \ge 0$. Then $a_n > 1$, $a_0 = 2$ and $$a_n = 1 + \sqrt{1 + a_{n-1} - 2\sqrt{a_{n-1}}} = \sqrt{a_{n-1}},$$ so $a_n = 2^{2^{-n}}$. Then $$\begin{align*} \sum_{n=1}^{N} b_n 2^n &= \sum_{n=1}^{N} (a_n - 1)^2 2^n = \sum_{n=1}^{N} \left[ a_n^2 2^n - a_n 2^{n+1} + 2^n \right] \\ &= \sum_{n=1}^{N} \left[ (a_{n-1} - 1) 2^n - (a_n - 1) 2^{n+1} \right] \\ &= (a_0 - 1) 2^1 - (a_N - 1) 2^{N+1} = 2 - 2\,\frac{2^{2^{-N}} - 1}{2^{-N}}. \end{align*}$$ Put $x = 2^{-N}$. Then $x \to 0$ as $N \to \infty$ and so $$\sum_{n=1}^{\infty} b_n 2^n = \lim_{N \to \infty} \left( 2 - 2\,\frac{2^{2^{-N}} - 1}{2^{-N}} \right) = \lim_{x \to 0} \left( 2 - 2\,\frac{2^x - 1}{x} \right) = 2 - 2 \ln 2.$$