An exponential series

Prove that

$\sum _{n=0}^{\infty} \frac {2^n}{3^{2^{n-1}} +1} = \frac{\sqrt{3}+1}{2}$

Solution

We recall the product identity

$\prod_{n=0}^{\infty} \left ( 1 + q^{2^n} \right ) = \frac{1}{1-q} \quad , \quad \left | q \right | < 1$

Thus,

$\begin{align*} \log \prod_{n=0}^{\infty} \left ( 1 + q^{2^n} \right ) = \log \frac{1}{1-q} &\Rightarrow \sum_{n=0}^{\infty} \log \left ( 1 + q^{2^n} \right )= -\log \left ( 1- q \right ) \\ &\Rightarrow \frac{\mathrm{d} }{\mathrm{d} q} \sum_{n=0}^{\infty} \log \left ( 1 + q^{2^n} \right ) = \frac{\mathrm{d} }{\mathrm{d} q} \left ( -\log (1-q) \right ) \\ &\Rightarrow \sum_{n=0}^{\infty} \frac{q^{2^n-1} 2^n}{q^{2^n}+1} = \frac{1}{1-q} \\ &\Rightarrow \sum_{n=0}^{\infty} \frac{q^{2^n}2^n}{q^{2^n}+1} = \frac{q}{1-q} \\ &\Rightarrow \sum_{n=0}^{\infty} \frac{2^n}{q^{-2^n} + 1} = \frac{q}{1-q} \\ &\!\!\!\!\!\!\!\!\!\!\overset{q=1/\sqrt{3}}{=\! =\! =\! =\! =\!\Rightarrow } \sum_{n=0}^{\infty} \frac{2^n}{3^{2^{n-1}} + 1} = \frac{\sqrt{3}+1}{2} \end{align*}$

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One thought on “An exponential series”

We can also get the value of the sum $\displaystyle \sum _{n=0}^{\infty} \frac {2^n}{3^{2^{n}} +1}$ . Indeed, if we consider $\displaystyle a_n = \frac{2^{n}}{3^{2^n} - 1}$ then we note that

$\begin{align*} a_n - a_{n+1} &= \frac{2^{n}}{3^{2^n} - 1} - \frac{2^{n+1}}{3^{2^{n+1}} - 1} \\ &= \frac{2^n}{3^{2^n} - 1} \left ( 1 - \frac{2}{3^{2^n} + 1} \right )\\ &= \frac{2^n}{3^{2^n} - 1} \frac{3^{2^n}-1}{3^{2^n}+1} \\ &= \frac{2^n}{3^{2^n}+1} \end{align*}$

and thus

$\begin{align*} \sum_{n=0}^{\infty} \frac{2^n}{3^{2^n}+1} &= \lim_{N \rightarrow +\infty} \sum_{n=1}^{N} \left ( a_n - a_{n+1} \right ) \\ &= a_0 - \lim_{N \rightarrow +\infty} a_{N+1} \\ &=\frac{1}{3^{2^0}-1} \\ &= \frac{1}{2} \end{align*}$

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One thought on “An exponential series”

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