On two identities of Ramanujan
Let \(|q|<1\), \[ (a;q)_\infty = \prod_{m=0}^\infty (1-aq^{m}), \] and \[ (a_1,a_2,\cdots,a_r;q)_\infty =(a_1;q)_\infty(a_2;q)_\infty\cdots (a_r;q)_\infty. \] Let \((\frac{\cdot}{p})\) denote the Legendre symbol modulo \(p\). By logarithmically differentiating both sides of \[ q^{-1}\frac{(q;q)_\infty^6}{(q^5;q^5)_\infty^6} = q^{-1}\frac{(q^2,q^3;q^5)^5_\infty}{(q,q^4;q^5)_\infty^5}-q \frac{(q,q^4;q^5)_\infty^5}{(q^2,q^3;q^5)^5_\infty}-11, \] and using the identity \[ 1-5\sum_{n=1}^\infty \left(\frac{n}{5}\right) \frac{nq^n}{1-q^n} =\frac{(q;q)_\infty^5}{(q^5;q^5)_\infty}, \] the authors prove Ramanujan's identity \[ 1+6\sum_{n=1}^\infty \frac{nq^n}{1-q^n} - 30\sum_{n=1}^\infty \frac{nq^{5n}}{1-q^{5n}} =\frac{(q;q)^5_\infty}{(q^5;q^5)_\infty}\left(1+22q\frac{(q^5;q^5)_\infty^6}{(q;q)_\infty^6}+125q^2\frac{(q^5;q^5)_\infty^{12}}{(q;q)_\infty^{12}}\right)^{1/2}. \] Using a similar method, they also prove that \[ 1+\sum_{k=1}^\infty \frac{kq^k}{1-q^k}-25\sum_{k=1}^\infty \frac{kq^{25k}}{1-q^{25k}} =q\frac{(q^5;q^5)_\infty^5}{(q;q)_\infty}\left(q^{-2}\frac{(q;q)_\infty^2}{(q^{25};q^{25})_\infty^2}+2q^{-1}\frac{(q;q)_\infty}{(q^{25};q^{25})_\infty}+5\right)^{1/2}. \]
scientific article; zbMATH DE number 1371469