calculate the $1/6+1/12+1/24+1/48 \ldots $. Wolfram is wrong? I am trying to calculate the following sum
$$
S = \frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48} + \cdots
$$
so
$$
S+\frac{1}{3} = \frac{1}{3} + \frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48} + \cdots = \frac{1}{3} \cdot \sum_{k=0}^{\infty} \frac{1}{2^k}
$$
This is equal to 
$$
\frac{1}{3} \cdot \frac{1}{1-0.5} = \frac{2}{3}
$$
so
$$
S + \frac{1}{3} = \frac{2}{3}
$$
and thus $S = \frac{1}{3}$.
So why in Wolfram Alpha when I do that [EDIT: new link ]
I get the following answer:
$$
0.3450320298895586027335724702689612099836962897927387
$$

You seem to have found a bug in Wolfram Alpha.  It appears to first interpret the sum $$\dfrac{1}{6} + \dfrac{1}{12} + \dfrac{1}{24} + \dfrac{1}{48} + \ldots $$ as $$\sum_{n=1}^\infty \dfrac{1}{n((n-3)n+8)} \approx 0.3450320299$$
(which does start $ \dfrac{1}{6} + \dfrac{1}{12} + \dfrac{1}{24} + \dfrac{1}{48}$, but the next term is $\dfrac{1}{90}$)
then after a few seconds "changes its mind" and writes the sum as
$$\sum_{n=1}^\infty \dfrac{1}{3 \times 2^n} $$
but leaves the rest of the answer unchanged.
