面白い級数1

$$\begin{array}{rcl}\left(左辺\right)& =& \sum _{k=1}^{\mathrm{\infty }}\frac{k{q}^k}{1+{(-q)}^k}\\ & & q(1+q+q^2+q^3+q^4+q^5+q^6+q^7+q^8+\cdots )\\ & +& 2q^2(1-q^2+q^4-q^6+q^8-\cdots )\\ & +& 3q^3(1+q^3+q^6+\cdots )\\ & +& 4q^4(1-q^4+\cdots )\\ & +& 5q^5(1+q^5+\cdots )\\ & +& \cdots \\ & =& \sum _{n=1}^{\mathrm{\infty }}{q}^n\sum _{d|n}d(-1{)}^{(d-1)\frac{n-d}{d}}\\ & =& \sum _{n=1}^{\mathrm{\infty }}{q}^n\sum _{d|n}\frac{n}{d}{(-1)}^{(d-1)\frac{n-d}{d}}\end{array}$$
$$\begin{array}{rcl}\left(右辺\right)& =& \sum _{k=1}^{\mathrm{\infty }}\frac{q^k}{{(1+{(-q)}^k)}^2)}\\ & & q(1+2q+3q^2+4q^3+5q^4+6q^5+\cdots )\\ & +& q^2(1-2q^2+3q^4-\cdots )\\ & +& q^3(1+2q^3+3q^6+\cdots )\\ & +& q^4(1-2q^4+3q^8-\cdots )\\ & +& q^5(1+2q^5+q^{10}+\cdots )\\ & +& \cdots \\ & =& \sum _{n=1}^{\mathrm{\infty }}{q}^n\sum _{d|n}\frac{n}{d}{(-1)}^{(d-1)\frac{n-d}{d}}\\ & =& \left(左辺\right)\end{array}$$

$$\begin{array}{r}\frac{m-n}{2^m}\frac{(m+n)!}{m!}=\frac{(m+n-1)!}{2^{m-1}(m-1)!}-\frac{(m+n)!}{2^{m}m!}\end{array}$$
より明らか。