$$\mathop {\lim }\limits_{n \to \infty } {{({1^a} + {2^a} + ... + {n^a})} \over {{{(n + 1)}^{a - 1}}[(na + 1) + (na + 2) + ... + (na + n)]}}$$

$$ = \mathop {\lim }\limits_{n \to \infty } {{\sum\limits_{r = 1}^n {{r^a}} } \over {{{(n + 1)}^{a - 1}}\left\{ {{n^2}a + {{n(n + 1)} \over 2}} \right\}}}$$

$$ = \mathop {\lim }\limits_{n \to \infty } {{{1 \over n}\sum\limits_{r = 1}^n {{{\left( {{r \over n}} \right)}^a}} } \over {{{\left( {1 + {1 \over n}} \right)}^{a - 1}}\left\{ {a + {1 \over 2} + {1 \over {2n}}} \right\}}}$$

$$ = {{\int\limits_a^1 {{x^a}dx} } \over {a + {1 \over 2}}} = {1 \over {(a + 1)\left( {a + {1 \over 2}} \right)}} = {1 \over {60}}$$ given.

$$ \Rightarrow (2a + 1)(a + 1) = 120 \Rightarrow 2{a^2} + 3a - 119 = 0$$

$$ \Rightarrow 2{a^2} - 14a + 17a - 119 = 0$$

$$ \Rightarrow 2a(a - 7) + 17(a - 7) = 0 \Rightarrow (2a + 17)(a - 7) = 0$$.

$$\therefore$$ $$a = 7,\, - {{17} \over 2}$$

But $$a = - {{17} \over 2}$$ has to be discarded because integral $$\int\limits_a^1 {{x^a}dx} $$ converges if a > $$-$$1