# A neat little sum

I was playing around with some infinite summations earlier today and I ended up discovering this little fact that kind of blew my mind a bit :)

$\frac{1}{n-1}+\frac{1}{n^2-1}+\frac{1}{n^3-1}+\cdots=\frac{d(1)}{n}+\frac{d(2)}{n^2}+\frac{d(3)}{n^3}+\cdots$

Assuming that the sums converge, and where the function $d(k)$ returns the number of divisors of $k$.

Proof: Note that each term on the left hand side is in the form $\frac{1}{n^{\alpha}-1}$. With a little bit of manipulation we can see that:

\begin{aligned} \frac{1}{n^{\alpha}-1} &= \frac{n^{-\alpha}}{n^{-\alpha}\left(n^{\alpha}-1\right)} \\ &= \frac{n^{-\alpha}}{1-n^{-\alpha}} \\ &= n^{-\alpha}+n^{-2\alpha}+n^{-3\alpha}+\cdots \end{aligned}

Returning to the original sum, we may now write it as:

$\begin{array}{c}&n^{-1} &+ &n^{-2} &+ &n^{-3} &+ &n^{-4} &+ &n^{-5} &+ &n^{-6} &+ &\cdots \\ &+ &n^{-2} & & &+ &n^{-4} & & &+ &n^{-6} &+ &\cdots \\ & & &+ &n^{-3} & & & & &+ &n^{-6} &+ &\cdots \\ & & & & &+ &n^{-4} & & & & &+ &\cdots \\ & & & & & & &+ &n^{-5} & & &+ &\cdots \\ & & & & & & & & &+ &n^{-6} &+ &\cdots \\ & & & & & & & & & & & \vdots & \ddots \end{array}$

With the first line representing the expansion of $\frac{1}{n-1}$, the second line representing $\frac{1}{n^2-1}$, etc. From this, we can see that the initial statement is true. We begin by summing $n^{-1}$ to the power of all multiples of $1$, then $n^{-1}$ to the power of all multiples of $2$, and so on. In other words, the number of divisors of the power of $n^{-1}$ tells us how many times it appears as a term in the whole expansion.

Has anyone else seen this result before? I'd like to read up on it or other results that are related to it!

Note by Daniel Hinds
2 years, 4 months ago

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Have a look at this mild generalization:

https://math.stackexchange.com/questions/273275/generating-function-for-the-divisor-function

- 2 years, 3 months ago