Some sum reciprocal

Note that:

$f(x) = \displaystyle \sum_{k=0}^\infty \frac {k+1}{x^k} = 1+2(\frac{1}{x})+3(\frac{1}{x})^2+4(\frac{1}{x})^3 \dots$

$x{f(x)} = x+2+3(\frac{1}{x})+4(\frac{1}{x})^2+5(\frac{1}{x})^3 \dots$

$x{f(x)}-f(x)=x+1+\frac{1}{x}+(\frac{1}{x})^2+(\frac{1}{x})^3 \dots$

$(x-1){f(x)} = x + \displaystyle \sum_{k=0}^\infty (\frac{1}{x})^k = x+\frac{1}{1-\frac{1}{x}} = x+\frac{x}{x-1} = \frac{x^2}{x-1}$

Therefore, $f(x) = \frac{\frac{x^2}{x-1}}{x-1} = \frac{x^2}{(x-1)^2}$

So, $f(-161) = \frac{(-161)^2}{(-161-1)^2} = \frac {{161}^2}{{162}^2} = S$

$f(x) = \frac{1}{S} = \frac {{162}^2}{{161}^2} = \frac{x^2}{(x-1)^2} \Rightarrow x=\boxed {162}$