Error Term in Passing from Summation to Integral I encountered the following in a paper and do not understand how the error term is being bounded. In what follows, $n$ and $k$ are large integer constants.

$$ \sum_{i=0}^{n-1} \ln\left(1 - \frac{i}{k}\right) = \int_0^n
 \ln\left(1 - \frac{x}{k}\right) dx \pm e(n,k) $$ where the error term
  $e(n,k)$ can be bounded by $$ e(n,k) \leq \int_0^n \ln\left(1 -
 \frac{x}{k}\right) - \ln\left(1 - \frac{x+1}{k}\right) dx  $$


$$0 \le e(n,k) = \int_0^n \left(\ln \left(1 - \dfrac{\lfloor x \rfloor}{k} \right) -  \ln \left(1 - \dfrac{x}{k}\right)\right) dx \le \int_0^n \left(\ln \left(1 - \dfrac{ x-1}{k} \right) -  \ln \left(1 - \dfrac{x}{k}\right)\right) dx $$
So the inequality is true if $$\ln \left(1 - \dfrac{ x-1}{k} \right) -  \ln \left(1 - \dfrac{x}{k}\right) \le \ln \left(1 - \dfrac{x}{k}\right) - \ln \left(1 - \dfrac{x+1}{k} \right)$$
but rearranging and taking exponential of both sides, this becomes
 $$\left(1 - \dfrac{ x-1}{k}\right)\left(1 - \dfrac{x+1}{k}\right) \le \left( 1 - \dfrac{x}{k}\right)^2$$
and the left side is $\left(1 - \dfrac{x}{k}\right)^2 - \left(\dfrac{1}{k}\right)^2$.
