# Deriving (a different) expression for the sum of divisors function.

I always really loved the derivation of the exact expression for the partition function, and I happened to recently stumble across this generating function for the sum of divisors function $\sigma(N)$:

$\hspace{7cm}\displaystyle\sum{\frac{Nz^N}{1-z^N}}=\sum{\sigma(N)z^N}$

I was wondering if it would be possible to use this generating function to derive some sort of analytic expression for $\sigma(N)$ in a similar way to how the exact formula for the partition function is derived.

I want to write something like:

$\hspace{6cm}\displaystyle \sigma(N)=\int_{\gamma}{\sum{\frac{Nz^N}{1-z^N}}e^{-2\pi iz}dz}$

and look at the asymptotic behavior of the integral for large values of N. Does anyone have any suggestions as to how to proceed?

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Note that the partition function depends only on the size of the argument, whereas the sum-of-divisors depends on the factorization; the partition function, so far as I know, is insensitive to primality or highly divisible nature of its argument. On the other hand, it means that the zeta function is available for analytic study of sum-of-divisors, and not for partitions. So my guess is you are barking up the wrong tree (but I'd be happy to be proven wrong). – Gerry Myerson Jan 15 '12 at 5:42
Not that I think this is likely to work, for reasons similar to what Gerry said, but you could start with the easier $\sum_N \frac{z^N}{1-z^N} = \sum_N \tau(N) z^N$, where $\tau$ is the number-of-divisors function. – Greg Martin Jan 15 '12 at 6:02

As a comment above indicates, the problem is that the sequence $\sigma(N)$ does not have an asymptotic behavior: as $N$ goes to infinity through the subsequence of primes, $\sigma(N)=N+1\sim N$. On the other hand, as $N$ goes to infinity through the subsequence of powers of $2$, $N=2^k$, $\sigma(N)=(2^{k+1}-1)/(2-1)$ which is $2\cdot 2^k-1\sim 2N$.