Question:
For \(a \in R\) (the set of all real numbers), \(a \ne - 1\), \[\mathop {\lim }\limits_{n \to \infty } \frac{{\left( {{1^a} + {2^a} + \cdots \cdot + {n^a}} \right)}}{{{{(n + 1)}^{a - 1}}[(na + 1) + (na + 2) + \cdots \cdot + (na + n)]}} = \frac{1}{{60}},\] then one of the values of \(a\) is
(a) 5
(b) 8
(c) \( - 15/2\)
(d) \( - 17/2\)

Answer:  \( - 17/2\)
Solution:

The function can be written as \[\mathop {\lim }\limits_{n \to \infty } \frac{{{n^a}\sum\limits_{r = 1}^n {{{\left( {\frac{r}{n}} \right)}^a}} }}{{{{(n + 1)}^{a - 1}}n\sum\limits_{r = 1}^n {\left( {a + \frac{r}{n}} \right)} }} = \frac{1}{{60}}\] \[ \Rightarrow \mathop {\lim }\limits_{n \to \infty } \,\,{\left( {\frac{n}{{n + 1}}} \right)^{a - 1}}\frac{{\sum\limits_{r = 1}^n {{{\left( {\frac{r}{n}} \right)}^a}} }}{{\sum\limits_{r = 1}^n {\left( {a + \frac{r}{n}} \right)} }}=\frac{1}{60}\] \[ = \mathop {\lim }\limits_{n \to \infty } \,\,\frac{{\sum\limits_{r = 1}^n {{{\left( {\frac{r}{n}} \right)}^a}} }}{{\sum\limits_{r = 1}^n {\left( {a + \frac{r}{n}} \right)} }} = \frac{{\int\limits_0^1 {{x^a}dx} }}{{\int\limits_0^1 {(a + x)dx} }}= \frac{2}{{(2a + 1)(a + 1)}}=\frac{1}{60}\] \[\Rightarrow a = 7,\; - \frac{{17}}{2}\]