Definite Integration Question - None

Question

If $\lim _\limits{n \rightarrow \infty} \frac{1^{a}+2^{a}+\ldots+n^{a}}{(n+1)^{a-1}[(n a+2)+\ldots(n a+n)]}=\frac{1}{60}$ for some positive real number a, then a is equal to:

Accepted Answer

$\lim _\limits{n \rightarrow \infty} \frac{\frac{1}{(a+1)} \cdot n^{a+1}+a_{1} n^{a}+a_{2} n^{a-1}+\ldots }{(n+1)^{a-1} \cdot n^{2}\left(a+\frac{1+\frac{1}{n}}{2}\right)}$ = $\frac 1{60}$
$\Rightarrow \lim _\limits{n \rightarrow \infty} \frac{\left(\frac{1}{n}\right)^{2}+\left(\frac{2}{n}\right)^{a}+\ldots +\left(\frac{n}{n}\right)^{a}}{(n+1)^{a-1}\left[n^{2} a+\frac{n(n+1)}{2}\right]}$ = $\frac 1{60}$
= $\frac{\lim _\limits{n \rightarrow \infty} \frac{1}{n} \sum_\limits{r=1}^{n}\left(\frac{r}{n}\right)^{a}}{\left(1+\frac{1}{n}\right)^{a-1}\left[a+\frac{1}{2}\left(1+\frac{1}{n}\right)\right]}$ = $\frac 1{60}$
= $\frac{\int_\limits{0}^{1} x^{a} d x}{\left(a+\frac{1}{2}\right)}=\frac{1}{60}=\frac{\frac{1}{a+1}}{a+\frac{1}{2}}$ = $\frac 1{60}$
= $\frac{\frac{1}{a+1}}{\left(a+\frac{1}{2}\right)}=\frac{1}{60}$
$\Rightarrow$ (a + 1)(2a + 1) = 120
$\Rightarrow$ 2a² + 3a - 119 = 0
$\Rightarrow$ 2a² + 17a - 14a - 119 = 0
$\Rightarrow$ (a - 7) (2a + 17) = 0
$\Rightarrow a=7,-\frac{17}{2}$

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