$(\tau(p)x-p^{11}x^2)\sum_{n=0}^{\infty}\tau(p^n)x^n=\sum_{n=0}^{\infty}\tau(p)\tau(p^n)x^{n+1}-\sum_{n=0}^{\infty}p^{11}\tau(p^n)x^{n+2}$
$=\sum_{n=0}^{\infty}\tau(p)\tau(p^n)x^{n+1}-\sum_{n=1}^{\infty}p^{11}\tau(p^{n-1})x^{n+1}$
$=\tau(p)x+\sum_{n=1}^{\infty}\tau(p)\tau(p^n)x^{n+1}-\sum_{n=1}^{\infty}p^{11}\tau(p^{n-1})x^{n+1}$
$=\tau(p)x+\sum_{n=1}^{\infty}\left(\tau(p)\tau(p^n)-p^{11}\tau(p^{n-1})\right)x^{n+1}$
$=\tau(p)x+\sum_{n=1}^{\infty}\tau(p^{n+1})x^{n+1}=\tau(p)x+\left(\sum_{n=-1}^{\infty}\tau(p^{n+1})x^{n+1} -\tau(p)x-1\right)$
$=\sum_{n=0}^{\infty}\tau(p^{n})x^{n} -1$ olur. Buradan $\sum_{n=0}^{\infty}\tau(p^n)x^n$ çözüldüğünde
$\displaystyle\sum_{n=0}^{\infty}\tau(p^n)x^n=\frac1{1-\tau(p)x+p^{11}x^2}$ elde edilir.