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:
$$
0.3450320298895586027335724702689612099836962897927387
$$

Here's a link. Wolfram Alpha looks right to me. Your link is rather unclear, likely truncated: what exactly did you look up on Alpha?
