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:

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

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

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

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!

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## Comments

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TopNewestHave a look at this mild generalization:

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

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