- Here we give another identity involving \(\theta ^ { 4 }\) , which is equivalent to the four-squares theorem.
- Show that for \(| q | < 1\)
\[ \sum_ {n = 1} ^ {\infty} \frac {n q ^ {n}}{1 - q ^ {n}} = \sum_ {n = 1} ^ {\infty} \frac {q ^ {n}}{(1 - q ^ {n}) ^ {2}}. \]
[Hint: The left-hand side is \(\sum \sigma _ { 1 } ( n ) q ^ { n }\) . Use \(\textstyle x / ( 1 - x ) ^ { 2 } = \sum _ { n = 1 } ^ { \infty } n x ^ { n } . ]\)
- Show as a result that
\[ \sum_ {n = 1} ^ {\infty} \frac {n q ^ {n}}{1 - q ^ {n}} - \sum_ {n = 1} ^ {\infty} \frac {4 n q ^ {4 n}}{1 - q ^ {4 n}} = \sum_ {n = 1} ^ {\infty} \frac {q ^ {n}}{(1 - q ^ {n}) ^ {2}} - 4 \sum_ {n = 1} ^ {\infty} \frac {q ^ {4 n}}{(1 - q ^ {4 n}) ^ {2}} = \sum \sigma_ {1} ^ {*} (n) q ^ {n} \]
where \(\sigma _ { 1 } ^ { * } ( n )\) is the sum of the divisors of n that are not divisible by 4.
- Show that the four-squares theorem is equivalent to the identity
\[ \theta (\tau) ^ {4} = 1 + 8 \sum_ {n = 1} ^ {\infty} \frac {q ^ {n}}{(1 + (- 1) ^ {n} q ^ {n}) ^ {2}}, \quad q = e ^ {\pi i \tau}. \]
Solution
For \(|q|<1\), all series below converge absolutely. Using \[ \frac{x}{(1-x)^2}=\sum_{k=1}^\infty kx^k, \] we get \[ \sum_{n=1}^\infty\frac{q^n}{(1-q^n)^2} =\sum_{n,k\ge1}kq^{nk}. \] Interchanging the dummy indices \(n\) and \(k\) gives \[ \sum_{n,k\ge1}nq^{nk} =\sum_{n=1}^\infty\frac{nq^n}{1-q^n}, \] which proves part (a).
Let \[ A(q)=\sum_{n=1}^\infty\frac{nq^n}{1-q^n} =\sum_{N=1}^\infty\sigma_1(N)q^N. \] Then \[ A(q)-4A(q^4) =\sum_{N=1}^\infty \bigl(\sigma_1(N)-4\sigma_1(N/4)\bigr)q^N, \] where \(\sigma_1(N/4)=0\) unless \(4\mid N\). The coefficient equals \[ \sum_{\substack{d\mid N\\4\nmid d}}d=:\sigma_1^*(N), \] because the divisors of \(N\) divisible by \(4\) are exactly \(4e\) with \(e\mid N/4\). This proves the first equality in part (b); part (a), applied at \(q\) and \(q^4\), proves the second.
Now put \[ B(q)=\sum_{n=1}^\infty \frac{q^n}{(1+(-1)^nq^n)^2}. \] Split into odd and even \(n\). For even \(n=2m\), \[ \frac{q^{2m}}{(1-q^{2m})^2} -4\frac{q^{4m}}{(1-q^{4m})^2} =\frac{q^{2m}}{(1+q^{2m})^2}. \] For odd \(n\), the summand is simply \(q^n/(1-q^n)^2\). Hence \[ B(q)=\sum_{n=1}^\infty\frac{q^n}{(1-q^n)^2} -4\sum_{n=1}^\infty\frac{q^{4n}}{(1-q^{4n})^2} =\sum_{N=1}^\infty\sigma_1^*(N)q^N. \]
Finally \[ \theta(\tau)^4 =\sum_{N=0}^\infty r_4(N)q^N, \qquad q=e^{\pi i\tau}. \] Thus Jacobi’s four-squares theorem \[ r_4(N)=8\sigma_1^*(N)\quad(N\ge1) \] is equivalent, coefficient by coefficient, to \[ \boxed{\theta(\tau)^4 =1+8\sum_{n=1}^\infty \frac{q^n}{(1+(-1)^nq^n)^2}}. \]