TITLE: Calculate the sum of inverse values of ${n\choose 0}, {n\choose 1}, ... {n\choose n}$
QUESTION [15 upvotes]: Calculate $$A={1\over {n\choose 0}}+ {1\over {n\choose 1}}+ ...+{1\over  {n\choose n}}$$
and
$$B={1\over {n\choose 0}}- {1\over {n\choose 1}}+ ...+{(-1)^n\over  {n\choose n}}$$ 

My idea for $A$ is some probabilistic reasoning. Color the sets $$\{\}, \{1\}, \{1,2\}, \{1,2,3\},...\{1,2,...,n\}$$
and ask our self what is the probability that I choose colored set among all sets. Clearly this is exactly ${n+1\over 2^n}$ and on the other side it is $A$:
probability that I take empty set is ${1\over {n\choose 0}}$, 
probability that I take colored set with 1 element is ${1\over {n\choose 1}}$
probability that I take colored set with 2 elements is ${1\over {n\choose 2}}$
and so on ... 
So $A ={n+1\over 2^n}$. But I have no idea how to attack $B$.

REPLY [5 votes]: I found the solution, though it is not mine. Write $${n\choose k}' := {(-1)^k\over {n\choose k}} $$
Then it is easy to prove $${n\choose k+1}'-{n\choose k}' =-{n+1\over n}{n-1\choose k}'$$
so we have:
\begin{eqnarray}
B&=&\sum _{k=0}^n {n\choose k}' \\&=&-{n+1\over n+2} \sum _{k=0}^n \Big[ {n+1\choose k+1}'-{n+1\choose k}'\Big] \\
&=& -{n+1\over n+2}\Big[ {n+1\choose n+1}'-{n+1\choose 0}'\Big] \\
&=& -{n+1\over n+2}\big( (-1)^{n+1}-1\big)\\
&=& {n+1\over n+2}\big( (-1)^n+1\big)\\
\end{eqnarray}
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