For a ∈ R (the set of all real numbers), a ≠ - 1 , lim n → ∞ 1 a + 2 a + … + n…

This numerical can be solved by using the concept of Definite integral as limit of sum.

$\underset{n\to \infty }{\text{Lim}}\frac{{\displaystyle \underset{r=1}{\overset{n}{\sum r^a}}}}{{\left(n+1\right)}^{a-1}\left[\sum _{r=1}^n(na+r)\right]}$

Divide Numerator and Denominator by $n^a$ and rearranging the terms

$\underset{n\to \infty }{\text{Lim}}\frac{{\displaystyle \sum _{r=1}^n{\left(\frac{r}{n}\right)}^a}}{{\displaystyle \left({\left(1+\frac{1}{n}\right)}^{a-1}\right)\left[\sum _{r=1}^n\left(a+\frac{r}{n}\right)\right]}}$

Now Multiply Numerator and denominator by $\frac{1}{n}$

$\underset{n\to \infty }{\text{Lim}}\frac{{\displaystyle \sum _{r=1}^n{\left(\frac{r}{n}\right)}^a\times \frac{1}{n}}}{{\displaystyle \left({\left(1+\frac{1}{n}\right)}^{a-1}\right)\left[\sum _{r=1}^n\left(a+\frac{r}{n}\right)\right]\times \frac{1}{n}}}$

Now Replace $\frac{1}{n}$ by $dx$and $\frac{r}{n}$ by $x$ and the lower limit and upper limit will be $\underset{n\to \infty }{\lim }\frac{1}{n}=0 ,\underset{n\to \infty }{\lim }\frac{n}{n}=1$,and $\sum _{}^{}$ will be replaced by ${\int }_{}^{}$

$$\begin{gathered} \frac{{\int }_{x=0}^1{x}^{a}dx}{{\int }_0^1\left(a+x\right)dx}=\frac{1}{60} \\ \frac{{\displaystyle {\left[{\displaystyle \frac{x^{a+1}}{a+1}}\right]}_0^1}}{{\displaystyle {\left[ax+{\displaystyle \frac{x^2}{2}}\right]}_0^1}}=\frac{1}{60} \\ \frac{{\displaystyle \frac{1}{a+1}-0}}{a+{\displaystyle \frac{1}{2}}-0}=\frac{1}{60} \\ \frac{2}{\left(a+1\right){\displaystyle \left(2a+1\right)}}=\frac{1}{60} \\ \left(a+1\right)\left(2a+1\right)=120 \\ 2a^2+3a+1=120 \\ 2a^2+3a-119=0 \\ a=\frac{-3\pm \sqrt{9+8\times 119}}{2\times 2} \\ a=\frac{-3\pm 31}{4} \\ a=\frac{28}{4} or -\frac{34}{4} \\ a=7 or -\frac{17}{2} \end{gathered}$$