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Does the convergent infinite sum $$ \sum_{n=0}^{\infty} \frac{1}{2^n + 1} $$ have a closed form solution? Quickly coding this up, the decimal approximation appears to be $1.26449978\ldots$

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Try this technique. – Mhenni Benghorbal Feb 4 '14 at 2:20
up vote 7 down vote accepted

Your series can be re-written in terms of the q-polygamma function $\psi_q(z)$ which is simply the logarithmic derivative of the q-gamma function $\Gamma_q(z)$. Both of which are special functions related to the theory of q-series: $$\sum_{n=0}^\infty\frac{1}{2^n+1}=\frac{\psi_{1/4}(1)-\psi_{1/2}(1)-\ln(3)}{\ln(2)}-\frac{3}{2}$$

Also as a consequence of several papers written by Erdos your sum is irrational. Erdos investigated similar series' when studying and also proving the irrationality of an analogous convergent series known as the "Erdős–Borwein constant" - the sum of the reciprocals of all the Mersenne numbers.

It also has several other series representations:

$$\sum_{n=0}^\infty\frac{1}{2^n+1}=\frac{1}{2}+\sum_{n=1}^\infty\frac{1}{2^n-1}-\sum_{n=1}^\infty\frac{2}{2^{2n}-1}=\frac{1}{2}-\sum_{n=1}^\infty\frac{(-1)^{n}}{2^n-1}=\frac{1}{2}-\sum_{n=1}^\infty\frac{\sum_{d\mid n}(-1)^d}{2^n}$$ $$=\frac{1}{2}+\sum_{n=1}^\infty\frac{1}{2^{n^2}}\frac{(2^n+1)}{(2^n-1)}-2\sum_{n=1}^\infty\frac{1}{4^{n^2}}\frac{(4^{n}+1)}{(4^{n}-1)}$$ $$=\frac{1}{2}+\sum_{n=1}^\infty\frac{1}{2^{n^2}}+2\sum_{n=1}^\infty\frac{1}{2^{n^2}(2^n-1)}-2\sum_{n=1}^\infty\frac{1}{4^{n^2}}-4\sum_{n=1}^\infty\frac{1}{4^{n^2}(4^{n}-1)}$$ $$=1+\frac{1}{2}\sum_{n=-\infty}^\infty\frac{1}{2^{n^2}}-\sum_{n=-\infty}^\infty\frac{1}{4^{n^2}}+2\sum_{n=1}^\infty\frac{1}{2^{n^2}(2^n-1)}-4\sum_{n=1}^\infty\frac{1}{4^{n^2}(4^{n}-1)}$$ $$=1+\frac{1}{2}\prod_{n=1}^\infty\frac{(2^{2n-1}+1)^2}{2^{6n-2}(2^{2n}-1)^{-1}}-\prod_{n=1}^\infty\frac{(2^{4n-2}+1)^2}{2^{12n-4}(2^{4n}-1)^{-1}}+\sum_{n=1}^\infty\frac{2}{2^{n^2}(2^n-1)}-\sum_{n=1}^\infty\frac{4}{4^{n^2}(4^{n}-1)}$$ Which comes from the Jacobi triple product identity.

Though in terms of computation, I would say the third series I gave which came as a result of a Lambert series identity would converge the quickest relative to the other expressions I listed:


Now in terms of a different "closed form" then the one at the top that uses the q-polygamma function. I would say that it's unlikely you're going to find another such representation in terms of anything other then a similar type of q-series based special function. Though similar lambert series' and q series expressions can take some very nice closed form values when their input is evaluated at exponentials of scaled values of $\pi$, like the series given in the link provided by Mhenni Benghorbal.

Also on a somewhat unrelated note, the inner partial sum appearing in the third series representation I gave for your sum: $$\sum_{d\mid n}(-1)^{d}=\frac{1}{2}(-1)^n\sum_{a^2-b^2=n}_{(a,b)\in \mathbb{Z}^2}1$$ Is equivalent to one half of negative one to the $n^{th}$ power multiplied by the number of representations of $n$ as a difference of the squares of two integers.

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First, you need to find the Mellin transform of the function

$$ \frac{1}{2^x+1}, $$

then just proceed with the technique in this problem.

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This wont work.. Unless you consider a closed form some unmanageable integral lol – Ethan Feb 4 '14 at 2:40
@Ethan: What do you mean exactly? Offcourse, you can find the Mellin transform of this function. – Mhenni Benghorbal Feb 4 '14 at 2:41
I mean the resulting Mellin transform wont be something clean that you can sum over and then invert again to get a nice expression, so far as I can tell. – Ethan Feb 4 '14 at 4:45

(Too long for comment)

For $n \ge 1$, $$ \frac{1}{1 + 2^n} = -\frac{(-1/2^n)}{1 - (-1/2^n)} = (-1/2^n) + (-1/2^n)^2 + (-1/2^n)^3 + (-2^n)^3 + \cdots $$

Which is an absolutely convergent series. Thus, \begin{align*} \sum_{n=1}^\infty \frac{1}{1 + 2^n} &= \sum_{n = 1}^\infty \sum_{m = 1}^\infty \frac{(-1)^m}{2^{mn}} \\ &= \sum_{N = 1}^\infty \sum_{d | N} \frac{(-1)^d}{2^N} \\ &= \sum_{N = 1}^\infty \frac{1}{2^N}\left[ \text{# of even divisors of } N- \text{# of odd divisors of } N\right] \\ \end{align*} Letting $N = 2^k l$ where $l$ is odd, and letting $\sigma_0$ be the number of divisors function, we get \begin{align*} \sum_{l \text{ odd }= 1}^\infty \sum_{k = 0}^\infty \frac{1}{2^{2^k l}} \left[ k\sigma_0(l) - \sigma_0(l)\right] &= \sum_{l \text{ odd }= 1}^\infty \sigma_0(l) \sum_{k = 0}^\infty (k-1) \left(\frac{1}{2^l}\right)^{2^k} \\ \end{align*}

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