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:
$$
0.3450320298895586027335724702689612099836962897927387
$$
 A: This turned out to be a look into WA's capabilities to do something useful with a human query.
If the queries are explicit, needing not too much inference by WA, we get a proper result:
Query with all information: sum 1/(6*2^k) for k from 0 to infty (link)
Here it has to guess the range: sum 1/(6*2^k) (link)  
Both give the correct result, but slightly different derivations.
For this one, which seems to be your original version, I get different results by whatever cluster node I might have hit: 1/6+1/12+1/24+1/48+ ... (link)
It requires WA to infer the general series from the few given terms.
E.g. I get a run where it assumes the general series
$$
\sum_{n=1}^\infty \frac{1}{n((n-3)n+8)} = 
\frac{1}{6} +
\frac{1}{12} + 
\frac{1}{24} + 
\frac{1}{48} + 
\frac{1}{90} + \ldots  
$$
which is what you seem to have gotten as well.
Note that this is not wrong, that general series hits the given terms, but deviates at the next not given term. However we generally mean the simplest answer and that means we would prefer 
$$
\sum_{n=0}^\infty\frac{1}{6\,2^k} \quad \mbox{or} \quad
\sum_{n=1}^\infty\frac{1}{3\,2^k} 
$$
which has the more complex term but is at least shorter (Kolmogorov-Chaitin complexity).
Here I made it a bit easier by factoring out the $1/6$: (1/6)(1+(1/2)+(1/4)+(1/8)+  ...) (link)
Interesting: Magic three dots v1 works in the WA iOS app:
 (Larger image version)
My impression is that via the app I get a bit more execution time on the WA cluster than when I enter via the web interface. So it has more recognition power. 
It looks like the engine tries several approaches to interpret a query until it hits the time limit and spews out the highest ranking result found so far.
A: 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?
A: 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.
