Closed form of $\sum\frac{1}{k}$ where $k$ has only factors of $2,3$ 
Consider the set containing $A$ all positive integers with no prime
  factor larger than $3$, and define $B$ as
  $$ 
B= \sum_{k\in A} \frac{1}{k} 
$$ 
  Thus, the first few terms of the sum are:
  $$
\frac{1}{1} +\frac{1}{2} +\frac{1}{3} +\frac{1}{4} +\frac{1}{6} +\frac{1}{8} +\frac{1}{9} +\frac{1}{12} +\frac{1}{16} +\frac{1}{18} + \cdots
$$
a) Write a
  closed-form expression for $X$ that makes the equation below true. In
  other words what expression should $X$ be so that the following
  equation is true, i.e. writing $B$ in terms of $X$. 
  $$
B = \sum_{i=0}^\infty \sum_{j=0}^\infty X
$$
  b) Write a closed-form expression for .


So there are obviously only two primes, $2$ and $3$. Thus, we have to consider the sum
$$\sum_{j\ge0}\sum_{i\ge0}\frac{1}{2^i3^j}$$
I would break this down further into the following:
$$\sum_{i\ge0}\frac{1}{3^0\cdot2^i}
+\sum_{i\ge0}\frac{1}{3^1\cdot2^i}
+\sum_{i\ge0}\frac{1}{3^2\cdot2^i}+\cdots$$
Now the first term is a geometric series with sum $\frac{1}{1-\frac{1}{2}}=2$. The second term is simply one third of that, or $\frac{2}{3}$. Continuing, the entire sum is equal to
$$\frac{2}{3^0}+\frac{2}{3^1}+\frac{2}{3^2}+\cdots=\frac{2}{1-\frac{1}{3}}=\boxed{3}$$
