An interesting sum

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

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

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

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

\(S = 2 (1 – \log 2)\).

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Ron Gordon

Math nerd in his early 40's who seems to have an opinion about everything and an inability to keep it to himself.