Answer
$ 1000$
Work Step by Step
Let $f(x)=\int_{1}^{\infty} \dfrac{1}{x^{1.001}} dx$
Now, $\lim\limits_{k \to \infty} f(x)= \lim\limits_{k \to \infty}\int_{1}^{k}\dfrac{1}{x^{1.001}} dx$
Thus, $\lim\limits_{k \to \infty} [\dfrac{-x^{1.001}}{(0.001)}]_{1}^{k}=\lim\limits_{k \to \infty} [-1000(k)^{-0.001}+1000]= 1000$
