Closed-Form Solution to Infinite Sum 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 $1.26449978\ldots$

(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*}
