Answer
converges to $\dfrac{1}{p-1}$ when $p \gt 1$
Work Step by Step
Here, $\int_1^n \dfrac{1}{x^p} dx=[\dfrac{1}{-p+1}(x^{-p+1}]_1^n$
and $a_n=\dfrac{1}{1-p}(\dfrac{1}{n^{(p-1)}-1})$
Let $\lim\limits_{n \to \infty} a_n=\lim\limits_{n \to \infty} (\dfrac{1}{1-p})(\dfrac{1}{n^{p-1}-1})$
Thus, $\dfrac{1}{n^{(p-1)}-1}$ approaches to $\infty$ when $p \lt 1$
and $\dfrac{1}{1-p}(\dfrac{1}{n^{(p-1)}-1}) \to \dfrac{1}{0}$ when $p=1$
Also, $p \gt 1$ so, $\dfrac{1}{1-p}(\dfrac{1}{n^{(p-1)}-1})$ approaches to $\dfrac{1}{(p-1)}$
Thus, {$a_n$} is Convergent and converges to $\dfrac{1}{p-1}$ when $p \gt 1$.
