Take the sum $\sum_1^{\infty} \frac{1}{2^n -1}$. I plugged it into Wolfram alpha, and it converges to a value around $1.606 \dots$ but wolfram didn't spit out any nice closed form. Is there an exact value for this sum?
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In general the sum
$$\sum_{k=1}^\infty\frac{1}{1-a^k}=\frac{\psi_{1/a}(1)+\ln(a-1)+\ln\frac 1 a}{\ln a}$$
where $\psi_q(z)$ is the $q$-digamma function.
Therefore
$$\sum _{k=1}^\infty\frac{1}{2^k-1}=1-\frac{\psi_{1/2}(1)}{\log 2}$$
Ben Longo
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No closed form (in terms of the familiar functions of high school algebra and intro Calculus) is known. The number is sometines called the Erdos-Borwein constant, e.g., here.
Gerry Myerson
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