## An interesting sum

Problem: compute, in closed form, the following sum:
$$S = \displaystyle\sum_{n=1}^{\infty} b_n 2^{n}$$,
where
$$b_n = 2 + \sqrt{b_{n-1}} – 2 \sqrt{1 + \sqrt{b_{n-1}}}$$,
and $$b_0 = 1$$.

Solution: Make the following substitution:

$$b_n = (p_n-1)^2$$.

Then we find that $$p_n^2 = p_{n-1}$$ and $$p_0 = 2$$. Therefore:
$$p_n = 2^{2^{-n}}$$.

The sum desired then takes the form

$$S = \displaystyle\sum_{n=1}^{\infty} \left (2^{2^{-n}} – 1 \right )^2 2^n$$.

It is not obvious how to go from here. It’s not even obvious from a cursory inspection that the sum converges. We can verify convergence by observing that

$$\displaystyle\left (2^{2^{-n}} – 1 \right )^2 2^n \sim (\log 2)^2 2^{-n} \; (n \to \infty )$$.

Now that we know that the sum converges, we can continue. We could expand the sum to evaluate, but there is a diverging piece which would give us fits (the $$2^n$$ piece). Knowing that the sum converges, and therefore the divergences cancel, we write

$$S = \displaystyle\lim_{m\to\infty} S_m$$
where
$$S_m = \displaystyle\sum_{n=1}^{m} \left (2^{2^{-n}} – 1 \right )^2 2^n$$.

The trick to see here is that
$$\displaystyle\left (2^{2^{-n}} \right)^2 = 2^{2^{-(n-1)}}$$.

Then
$$S_m = \displaystyle\sum_{n=1}^{m}2^{2^{-(n-1)}} 2^{n} – 2 \displaystyle\sum_{n=1}^{m}2^{2^{-n}} 2^n + \displaystyle\sum_{n=1}^{m} 2^n$$, or

$$S_m = 2 \displaystyle\sum_{n=0}^{m-1}2^{2^{-n}} 2^{n} – 2 \displaystyle\sum_{n=1}^{m}2^{2^{-n}} 2^n + 2^{m+1}-2$$.

The terms in the first two sums all cancel except for the first term in the first sum and the last term in the last sum. We can then write $$S_m$$ in closed form:

$$S_m = 2 – 2 \displaystyle\left (2^{2^{-m}} – 1 \right ) 2^m$$.

Using the fact, recited above, that

$$\displaystyle\lim_{m\to\infty}\left (2^{2^{-m}} – 1 \right ) 2^m = \log 2$$,

we can then write the solution as

$$S = 2 (1 – \log 2)$$.