For a ∈ R (the set of all real numbers), a ≠ - 1 , _ n → ∞ ( 1 ^ a...

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

$\lim_{n \to \infty}\dfrac{\mathop{∑ {r}^{a}}_{r = 1}^{n}}{{\left(n + 1\right)}^{a - 1}\left[\mathop{∑}_{r = 1}^{n} ( na + r )\right]}$

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

$\lim_{n \to \infty}\dfrac{\mathop{∑}_{r = 1}^{n}{\left(\dfrac{r}{n}\right)}^{a}}{\left({\left(1 + \dfrac{1}{n}\right)}^{a - 1}\right)\left[\mathop{∑}_{r = 1}^{n}\left(a + \dfrac{r}{n}\right)\right]}$

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

$\lim_{n \to \infty}\dfrac{\mathop{∑}_{r = 1}^{n}{\left(\dfrac{r}{n}\right)}^{a} \times \dfrac{1}{n}}{\left({\left(1 + \dfrac{1}{n}\right)}^{a - 1}\right)\left[\mathop{∑}_{r = 1}^{n}\left(a + \dfrac{r}{n}\right)\right] \times \dfrac{1}{n}}$

Now Replace $\dfrac{1}{n}$ by $dx$and $\dfrac{r}{n}$ by $x$ and the lower limit and upper limit will be $\lim_{n \to \infty}\dfrac{1}{n} = 0, \lim_{n \to \infty}\dfrac{n}{n} = 1$,and $\mathop{∑}_{}^{}$ will be replaced by ${∫}_{}^{}$

$\dfrac{{∫}_{x = 0}^{1}{x}^{a} d x}{{∫}_{0}^{1}\left(a + x\right) d x} = \dfrac{1}{60}\dfrac{{\left[\dfrac{{x}^{a + 1}}{a + 1}\right]}_{0}^{1}}{{\left[ax + \dfrac{{x}^{2}}{2}\right]}_{0}^{1}} = \dfrac{1}{60}\dfrac{\dfrac{1}{a + 1} - 0}{a + \dfrac{1}{2} - 0} = \dfrac{1}{60}\dfrac{2}{\left(a + 1\right)\left(2a + 1\right)} = \dfrac{1}{60}\left(a + 1\right)\left(2a + 1\right) = 1202{a}^{2} + 3a + 1 = 1202{a}^{2} + 3a - 119 = 0a = \dfrac{- 3 \pm \sqrt{9 + 8 \times 119}}{2 \times 2}a = \dfrac{- 3 \pm 31}{4}a = \dfrac{28}{4}or - \dfrac{34}{4}a = 7or - \dfrac{17}{2}$