Compute $\sum\arctan\frac1{3^n}$ Find the Closed-form expression of the sum
$$
\sum_{n=1}^{\infty}\arctan\left(1 \over 3^{n}\right)
$$
Notice that $\arctan\left(x\right) < x,\forall\ x > 0$, so
$\sum_{n = 1}^{\infty}\arctan\left(1 \over 3^{n}\right) < {1 \over 2}$.
 A: Notice for any $x \in \mathbb{R}$, we have
$$\tan^{-1}x = \Im\log (1 + ix)$$
This implies
$$\sum_{n=1}^\infty \tan^{-1}\frac{1}{3^n} 
= \Im\left\{\sum_{n=1}^\infty \log\left( 1 + \frac{i}{3^n}\right)\right\}
= \Im\left\{ \log \prod_{n=1}^\infty \left(1 + \frac{i}{3^n}\right)\right\} + 2\pi N
$$
for some $N \in \mathbb{Z}$. Please note that $\log z$ is not well defined over the whole
$\mathbb{C}$, it is defined only up to an integer multiple of $2\pi i$. That's why we convert the sum of log to log of product, we will pick up an extra factor $2\pi N$. However, the sum in LHS is small enough and falls within the range $(-\frac{\pi}{2},\frac{\pi}{2})$, so $N = 0$ in this particular case.
The q-Pochhammer symbol $(a;q)_{n}$
where $n \in \mathbb{N} \cup \{ \infty \}$ is defined as
$$( a; q )_n = \prod_{k=0}^n (1 - aq^k )$$
Comparing with the equality above, we get
$$\sum_{n=1}^\infty \tan^{-1}\frac{1}{3^n} = \Im\log\left( -\frac{i}{3}, \frac13 \right)_\infty$$
On WA, we can evaluate RHS using the expression Im[Log[QPochhammer[-i/3,1/3]]] 
and it gives us an value
$$\approx 0.487945758671045507547332675668307675519350963398192969940241$$
In certain sense, this is just a recast of the original series to a more or less
equivalent known function. The only advantage of this is the q-Pochhammer symbols are well studied. There are more efficient algorithms to compute its value instead of preforming the sum of $\tan^{-1}(\cdots)$ ourselves.
A: Since user92774 gave you an approximate value, if you use the Taylor expansion of $\arctan(x)$, the result of your summation (don't forget that the argument is is geometric progression) is    
$1/2-1/78 + 1/1210 - 1/15302 + 1/177138 - 1/1948606 + 1/20726186 - 1/215233590 + ...$   
The summation of these few terms gives $0.487945758256695$ while the $100,000$-th partial sum is $0.487945758671045$.
