Newton's method for finding zeroes of a function uses an iterative step like this:
\begin{align*} x_{k+1} = x_k - \frac{f(x_k)}{f'(x_k)} \\ \end{align*}
The following variation will instead find poles of a function. why? \begin{align*} x_{k+1} = x_k + \frac{f(x_k)}{f'(x_k)} \\ \end{align*}
Taking the limit as $k\to\infty$ in both expressions gives $f(x_k)/f'(x_k)\to 0$. When searching for zeros, it is natural that $f(x_k)\to 0$. However, when searching for poles, it is not, but $f'(x_k)\to \pm\infty$ is. Since searching the poles of $f$ is equivalent to searching the zeros of $1/f$, we have \begin{aligned} x_{k+1} &= x_k - \frac{1/f(x_k)}{-f'(x_k)/f(x_k)^2} \, ,\\ &= x_k + \frac{f(x_k)}{f'(x_k)} \, . \end{aligned} The simplification by $f(x_k)^2$ is responsible for the confusion.