I am trying to prove that the sequence $I_n(p)=\int_1^n \frac{dx}{x^p}$ does or does not converge uniformly to the function $I(p)=\int_1^\infty \frac{dx}{x^p}$ for $p>1$.
It seems to me that $\sup |I_n(p)-I(p)|= \lim_{p \to 1} (I(p)-I_n(p))$. Is it enough, then, to show that for all $\epsilon>0$ there exists a natural number $N$ such that for $p\to 1^+$ and all $n\geq N$, $I(p)-I_n(p) < \epsilon$? Can I set $p = 1$?
Edit:
$I(p)-I_n(p) = \int_1^\infty \frac{dx}{x^p} - \int_1^n \frac{dx}{x^p} = \int_n^\infty \frac{dx}{x^p}$.
$\lim_{n\to\infty} \lim_{p\to 1} \int_n^\infty \frac{dx}{x^p} = \lim_{n\to\infty} \int_n^\infty \frac{dx}{x} = 0$
Is this correct?