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I hope this is not a duplicate question. If we modify the well known geometric series, with $a>1$, to

$$ \sum_{n=0}^{\infty} \frac{1}{1+a^n} $$ is there a closed form with a name?

I suspect strongly that the answer is not in terms of elementary functions but will be some special function defined as such (as e.g. of the likes of the Hurwitz zeta function)

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  • $\begingroup$ Maple, Mathematica (and others) will give exotic "closed forms" that are essentially tautologous. $\endgroup$ – André Nicolas Jul 29 '15 at 3:58
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    $\begingroup$ Replacing $a$ with $1/a$, discarding the $n=0$ term and expanding yields OEIS A048272 $\endgroup$ – ccorn Jul 29 '15 at 4:01
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By expanding $\frac{1}{1+a^n}$ as a geometric series,

$$\sum_{n\geq 1}\frac{1}{1+a^n}=\sum_{n\geq 1}\sum_{m\geq 1}\frac{(-1)^{m+1}}{a^{mn}}=\sum_{l\geq 1}\frac{1}{a^l}\sum_{d\mid l}(-1)^{\frac{l}{d}+1}=\sum_{l\geq 1}\frac{g(l)}{a^l}$$ where $g(l)$ is an arithmetic function counting the difference between the number of odd divisors and the number of even divisors. $-g$ is a multiplicative function, for which: $$ -g(2^k) = 1-k,\qquad -g(p^k) = 1+k,$$ hence: $$\sum_{n\geq 1}\frac{1}{1+a^n}=\sum_{\nu=0}^{+\infty}(\nu-1)\sum_{q=0}^{+\infty}\frac{d(2q+1)}{a^{2^\nu(2q+1)}}.$$

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We can write $$\frac{1}{1+a^n}=\left[\frac{\partial}{\partial t}\ln\left(1+ta^{-n}\right)\right]_{t=1},$$ which implies that $$\sum_{n=0}^{\infty}\frac{1}{1+a^n}=\left[\frac{\partial}{\partial t}\ln\prod_{n=0}^{\infty}\left(1+ta^{-n}\right)\right]_{t=1}= \left[\frac{\partial}{\partial t}\,\ln\left(-t;a^{-1}\right)_{\infty}\right]_{t=1},$$ where $(z,q)_{\infty}$ denotes the q-Pochhammer symbol.

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