If lim n → ∞ ⁡ 1 a + 2 a + … + n a n + 1 a - 1 n a + 1 + n a + 2 + … + n a + n…

$\underset{n\to \infty }{\lim }\frac{n^a\left({\displaystyle \sum _{r=1}^n{\left(\frac{r}{n}\right)}^a}\right)}{{(n+1)}^{a-1}.n^2(a+\frac{1+\frac{1}{n}}{2})}=\frac{1}{60}$

$\Rightarrow \underset{n\to \infty }{\lim }\frac{n^a{\displaystyle \sum _{r=1}^{}}{\left({\displaystyle \frac{r}{n}}\right)}^a}{{\left(n\right)}^{a+1}{\left(1+{\displaystyle \frac{1}{n}}\right)}^{a-1}\left(a+{\displaystyle \frac{1+{\displaystyle \frac{1}{n}}}{2}}\right)}=\frac{1}{60}$

$\Rightarrow \underset{n\to \infty }{\lim }\frac{{\displaystyle \frac{1}{n}\sum _{r=1}^n{\left(\frac{r}{n}\right)}^a}}{{(1+\frac{1}{n})}^{a-1}{\displaystyle \left(a+\frac{1+{\displaystyle \frac{1}{n}}}{2}\right)}}=\frac{1}{60}$

$\Rightarrow \frac{{\displaystyle \underset{0}{\overset{1}{\int }}{x}^{a}dx}}{(a+\frac{1}{2})}=\frac{1}{60}$

$\Rightarrow \frac{\frac{1}{a+1}}{\left(a+\frac{1}{2}\right)}=\frac{1}{60}$

$\Rightarrow \left(a+1\right) \left(2a+1\right)=120$

$\Rightarrow 2a^2+3a-119=0$

$\Rightarrow 2a^2+17a-14a-119=0$

$\Rightarrow \left(a-7\right) \left(2a+17\right)=0$

$a=7, -\frac{17}{2}$.