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 returns the number of divisors of .
Proof: Note that each term on the left hand side is in the form . With a little bit of manipulation we can see that:
Returning to the original sum, we may now write it as:
With the first line representing the expansion of , the second line representing , etc. From this, we can see that the initial statement is true. We begin by summing to the power of all multiples of , then to the power of all multiples of , and so on. In other words, the number of divisors of the power of 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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Have a look at this mild generalization:
https://math.stackexchange.com/questions/273275/generating-function-for-the-divisor-function