Sum of reciprocals of prime factors

Question

What is the value of following sum?

$$\sum_{n_1,n_2,n_3 =0} ^ \infty \frac{1}{2^{n_{1}} 3^{n_{2}} 5^{n_{3}} } $$

Where $n_1,n_2,n_3$ are positive integers.

Answer 1

Write the sum as: $$\sum_{n_1=0}^\infty \left(\sum_{n_2=0}^\infty \left(\sum_{n_3=0}^\infty \frac{1}{2^{n_1}3^{n_2}5^{n_3}}\right)\right).$$ In the innermost sum, $2^{n_1}$ and $3^{n_2}$ are constant, so we can factor them out: $$\sum_{n_1=0}^\infty \left(\sum_{n_2=0}^\infty \frac{1}{2^{n_1}3^{n_2}} \left(\sum_{n_3=0}^\infty \frac{1}{5^{n_3}}\right)\right).$$ Again, $2^{n_1}$ is a constant in the middle sum, so we can factor it out: $$\sum_{n_1=0}^\infty \frac{1}{2^{n_1}}\left(\sum_{n_2=0}^\infty \frac{1}{3^{n_2}}\left(\sum_{n_3=0}^\infty \frac{1}{5^{n_3}}\right)\right).$$ Now whatever $\sum_{n_3}^\infty \frac{1}{5^{n_3}}$ is (you can work it out using the formula for the sum of a geometric series), it is constant with respect to $n_1$ and $n_2$, so it can be factored out: $$\left(\sum_{n_3=0}^\infty \frac{1}{5^{n_3}}\right)\left(\sum_{n_1=0}^\infty \frac{1}{2^{n_1}} \left(\sum_{n_2=0}^\infty \frac{1}{3^{n_2}}\right)\right).$$ Similarly, whatever $\sum_{n_2=0}^\infty \frac{1}{3^{n_2}}$ is, it is constant with respect to $n_1$ and can be factored out of the sum: $$\left(\sum_{n_3=0}^\infty \frac{1}{5^{n_3}}\right)\left(\sum_{n_2=0}^\infty \frac{1}{3^{n_2}}\right)\left(\sum_{n_1=0}^\infty \frac{1}{2^{n_1}}\right).$$ Now use the formula for the sum of a geometric series.