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<math> \text{Ave}_{n}\sigma_{k}=\frac{1-q^{k(n+1)}}{1-q^{k}}</math>.
Notice that
<math>q^{n}Ave_{n}\sigma_{k}=q^{n(k+1)}(\frac{1-q^{-k(n+1)}}{1-q^{-k}})=q^{n(k+1)}(\frac{\zeta(k+1)}{\zeta(kn+k+1)})</math>
Thus, if we set <math>x=q^{n}</math> then the above result reads
<math>\sum_{\text{deg}(m)=n, m \text{ monic}} \sigma_k(m)=x^{k+1}(\frac{\zeta(k+1)}{\zeta(kn+k+1)})</math>,
which resembles closely the analogous result for the integers:
====Number of divisors====
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