Closed form of $\sum\frac{1}{k}$ where $k$ has only factors of $2,3$

Question

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 .

Answer 1

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

$$\frac{2}{3^0}+\frac{2}{3^1}+\frac{2}{3^2}+\cdots=\frac{2}{1-\frac{1}{3}}=\boxed{3}$$

Answer 2

In general, you use the same logic used for the Euler product. Notice that your form becomes $$(1+\sum\limits_{k=1}^{\infty}\frac{1}{2^k})(1+\sum\limits_{k=1}^{\infty}\frac{1}{3^k})$$ since you have all powers of 2 and all powers of 3 and when you multiply them you have all possible combinations without repetition, and since prime decomposition is unique we are sure we have the same series as you have required. But this is

$$(1+\sum\limits_{k=1}^{\infty}\frac{1}{2^k})(1+\sum\limits_{k=1}^{\infty}\frac{1}{3^k})=\frac{1}{1-\frac{1}{2}}\frac{1}{1-\frac{1}{3}}=3$$

If you ask the same question for other cases and you know the prime numbers involved $p_{1},p_{2},...,p_{m}$ you will have the answer as

$$\prod_{k=1}^{m}\frac{p_{k}}{p_{k}-1}$$