For $a \in \mathbb{R}$ (the set of all real numbers), $a \neq -1$, $\lim_{n \to \infty} \frac{(1^{a} + 2^{a} + \dots + n^{a})}{(n+1)^{a-1}[(na+1) + (na+2) + \dots + (na+n)]} = \frac{1}{60}.$ Then $a =$

For a ∈ R (the set of all real numbers), a ≠ -1, lim_(n → ∞) ((1ᵃ + 2ᵃ + … + nᵃ))/((n+1)ᵃ⁻¹[(na+1) + (na+2) + … + (na+n)]) = 1/60. Then a =

$\lim_{n \to \infty} \dfrac{1^{a} + 2^{a} + \dots + n^{a}}{(n+1)^{a-1}[(na+1) + \dots + (na+n)]} = \dfrac{1}{60}$

$\text{To find: } a$

$A_n=\sum_{k=1}^{n}k^{a},\qquad D_n=(n+1)^{a-1}\sum_{k=1}^{n}(na+k)$

$\sum_{k=1}^{n}(na+k)=n\cdot na+\dfrac{n(n+1)}{2}=n^2a+\dfrac{n(n+1)}{2}$

$= n^2\left(a+\dfrac{n+1}{2n}\right)=n^2\left(a+\dfrac{1}{2}+\dfrac{1}{2n}\right)$

$\text{Plan: } A_n \sim n^{a+1}\!\int_0^1 x^a\,dx;\quad \text{form the ratio} \Rightarrow a$

$A_n=n^{a+1}\cdot\dfrac{1}{n}\sum_{k=1}^{n}\left(\dfrac{k}{n}\right)^{a}$

$\dfrac{1}{n}\sum_{k=1}^{n}\left(\dfrac{k}{n}\right)^{a}\to\int_0^1 x^a\,dx=\dfrac{1}{a+1}$

$\Rightarrow A_n \sim \dfrac{n^{a+1}}{a+1}$

$(n+1)^{a-1}\sim n^{a-1} \Rightarrow D_n \sim n^{a-1}\cdot n^2\left(a+\dfrac{1}{2}\right)=n^{a+1}\left(a+\dfrac{1}{2}\right)$

$\lim_{n\to\infty}\dfrac{A_n}{D_n}=\dfrac{1}{(a+1)\left(a+\dfrac{1}{2}\right)}$

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

$a^2+\dfrac{3}{2}a+\dfrac{1}{2}=60 \;\Rightarrow\; 2a^2+3a-119=0$

$a=\dfrac{-3\pm\sqrt{9+952}}{4}=\dfrac{-3\pm 31}{4} \;\Rightarrow\; a=7 \ \text{ or } \ a=-\dfrac{17}{2}$

$A_n\sim\dfrac{n^{a+1}}{a+1}\ \text{ requires } a>-1$

$-\dfrac{17}{2}<-1 \;\Rightarrow\; \text{discard},\qquad a=7$

$(7+1)\left(7+\dfrac{1}{2}\right)=8\cdot\dfrac{15}{2}=60\ \checkmark$

$a=7 \quad \text{option }(2)$