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)d(k) returns the number of divisors of kk.

Proof: Note that each term on the left hand side is in the form 1nα1\frac{1}{n^{\alpha}-1}. With a little bit of manipulation we can see that:

1nα1=nαnα(nα1)=nα1nα=nα+n2α+n3α+\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:

n1+n2+n3+n4+n5+n6++n2+n4+n6++n3+n6++n4++n5++n6+\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 1n1\frac{1}{n-1}, the second line representing 1n21\frac{1}{n^2-1}, etc. From this, we can see that the initial statement is true. We begin by summing n1n^{-1} to the power of all multiples of 11, then n1n^{-1} to the power of all multiples of 22, and so on. In other words, the number of divisors of the power of n1n^{-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!

#NumberTheory

Note by Daniel Hinds
2 years, 2 months ago

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Comments

Have a look at this mild generalization:

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

Patrick Corn - 2 years, 2 months ago
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