- Recall from Problem 2 in Chapter 2, that
\[ \sum_ {n = 1} ^ {\infty} d (n) z ^ {n} = \sum_ {n = 1} ^ {\infty} \frac {z ^ {n}}{1 - z ^ {n}}, \quad | z | < 1 \]
where \(d ( n )\) denotes the number of divisors of \(n\) .
More generally, show that
\[ \sum_ {n = 1} ^ {\infty} \sigma_ {\ell} (n) z ^ {n} = \sum_ {n = 1} ^ {\infty} \frac {n ^ {\ell} z ^ {n}}{1 - z ^ {n}}, \quad | z | < 1 \]
where \(\sigma _ { \ell } ( n )\) is the sum of the \(\ell ^ { \mathrm { t h } }\) powers of divisors of \(n .\) .
Solution
Fix \(\abs z<1\) and put \(r=\abs z\).
1. For every \(n\ge1\), \(\dfrac{n^\ell z^n}{1 - z^n} = \displaystyle\sum_{k=1}^\infty n^\ell z^{nk}\).
Since \(\abs{z^n} = r^n < 1\), the geometric series gives \(\frac{z^n}{1 - z^n} = \sum_{k\ge1} z^{nk}\).
2. The double series \(\sum_{n\ge1} \sum_{k\ge1} n^\ell z^{nk}\) converges absolutely.
Using \(\frac{r^n}{1 - r^n} \le \frac{r^n}{1 - r}\), \[\sum_{n=1}^\infty \sum_{k=1}^\infty n^\ell r^{nk} = \sum_{n=1}^\infty n^\ell \frac{r^n}{1 - r^n} \le \frac{1}{1 - r} \sum_{n=1}^\infty n^\ell r^n,\] and \(\sum n^\ell r^n\) converges by the ratio test.
3. Q.E.D.
By step 2 the double series may be summed along the values \(m = nk\). By step 1, \[\sum_{n=1}^\infty \frac{n^\ell z^n}{1 - z^n} = \sum_{n=1}^\infty \sum_{k=1}^\infty n^\ell z^{nk} = \sum_{m=1}^\infty \Bigl( \sum_{n \mid m} n^\ell \Bigr) z^m = \sum_{m=1}^\infty \sigma_\ell(m) z^m.\]