# Sum of reciprocals of prime factors

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.

• I don't understand the notation, could you please expand the sum? – Dr. Mathva May 10 at 16:27
• It is $\frac{1}{1-2^{-1}}\frac{1}{1-3^{-1}}\frac{1}{1-5^{-1}}$ – reuns May 10 at 16:33
• $3.75{}{}{}{}{}$ – Saucy O'Path May 10 at 16:33
• Hint: $\sum_{i,j= 0}^\infty a_ib_j = (\sum_{i=0}^\infty a_i)(\sum_{j=0}^\infty b_j)$. Do you see why this is? – eyeballfrog May 10 at 16:33

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.