Series of positive factors of a number divided by that number

Question

Let $S_n$ be the sum of the positive factors of $2015^n$, with $n$ being a positive integer approaching infinity. What is $\dfrac{S_n}{2015^n}$?

I might be on the wrong track, but I figure that if $x = \frac 15$, $y = \frac 1{13}$, and $z = \frac 1{31}$ (the reciprocals of the prime factors of $2015$), every term in the series must be some combination of those: $x^0y^0z^0$ + $x^1y^0z^0$ + $x^0y^1z^0$ and so on using every combination of exponents up to $x^ny^nz^n$. At first I was thinking to figure out a polynomial expansion then I'd have a converging infinite geometric series, but I couldn't for the life of me find the right way to get it to make that series (or if it's even possible using that method). I know it is pretty darn close to $1.4$, but I was hoping for something with a little more rigor. Any help would be great.

Answer 1

You had the right intuition that $S_n$ would be some sort of geometric series.

Note that the sum of the positive factors of $2015^n$ is given by $$S_n = (1+5+5^2+\cdots+5^n)(1+13+13^2+\cdots+13^n)(1+31+31^2+\cdots+31^n).$$ Do you see why? (If you don't, try FOILing that product and see what terms are being added).

From this, it follows that $\dfrac{S_n}{2015^n} = \dfrac{S_n}{5^n \cdot 13^n \cdot 31^n} = $

$(1+5^{-1}+5^{-2}+\cdots+5^{-n})(1+13^{-1}+13^{-2}+\cdots+13^{-n})(1+31^{-1}+31^{-2}+\cdots+31^{-n})$.

By taking the limit as $n \to \infty$, you get the product of three infinite geometric series, each of which is easy to compute.

Answer 2

HINT: If $n =\prod_{k} p_k^{\alpha_k}$, then sum of all divisor of $n$ is $$\prod_{k} \frac{p_k^{\alpha_k+1}-1}{p_k-1}$$.

Now, note that $2015 = 5. 13. 31$, then you get $$S_n = \frac{5^{n+1}-1}{5-1}\frac{13^{n+1}-1}{13-1}\frac{31^{n+1}-1}{31-1} = \frac{1}{1440} (5^{n+1}-1)(13^{n+1}-1)(31^{n+1}-1)$$.