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Numerically, they seem to be the same, but can we prove that $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n\frac{\sigma(k)}{k}=\zeta(2),$$ where $\sigma(k)$ is the sum of divisors of $k$.

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up vote 11 down vote accepted

Notice that $\frac{\sigma(k)}{k} = \displaystyle \sum_{d | k} \frac{1}{d}$, since $\sigma(k)$ is just the sum of the divisors of $k$.

Then we can write $$ \frac{1}{n} \sum_{k=1}^n \frac{\sigma(k)}{k} = \frac{1}{n} \sum_{k=1}^n \left(\sum_{d | k} \frac{1}{d}\right) = \frac{1}{n} \sum_{d=1}^n \frac{\left[\frac{n}{d}\right]}{d} $$ since the term $\frac{1}{d}$ shows up once for each $k$ from 1 to $n$ that is a multiple of $d$.

Taking the limit as $n \rightarrow \infty$, we get $$ \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{d=1}^n \frac{\left[\frac{n}{d}\right]}{d} = \sum_{d=1}^\infty \frac{1}{d^2} = \zeta(2). $$

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Aah. I was thinking that this might be a numerical coincidence, but this is easy to generalize to get any $\zeta(m)$. Thank you for the very nice answer! – Carolus Apr 18 '12 at 7:27

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