If mathop {lim }limits_{n to infty } frac{{{1 | vedclass

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(A) The given limit is $\mathop {\lim }\limits_{n 	o \infty } \frac{{\sum_{r=1}^n r^a}}{{(n+1)^{a-1} \sum_{r=1}^n (na + r)}} = \frac{1}{60}$.Divide t

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$7$

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(A) The given limit is $\mathop {\lim }\limits_{n 	o \infty } \frac{{\sum_{r=1}^n r^a}}{{(n+1)^{a-1} \sum_{r=1}^n (na + r)}} = \frac{1}{60}$.Divide the numerator and denominator by $n^{a+1}$:$\mathop {\lim }\limits_{n 	o \infty } \frac{{\frac{1}{n} \sum_{r=1}^n (\frac{r}{n})^a}}{{(\frac{n+1}{n})^{a-1} \cdot \frac{1}{n^2} \sum_{r=1}^n (na + r)}} = \frac{1}{60}$.Using the definition of definite integral $\int_0^1 x^a dx = \frac{1}{a+1}$ and the sum of arithmetic progression $\sum_{r=1}^n (na + r) = n^2 a + \frac{n(n+1)}{2}$,we get:$\frac{\int_0^1 x^a dx}{\lim_{n 	o \infty} (1 + \frac{1}{n})^{a-1} (a + \frac{1}{2}(1 + \frac{1}{n}))} = \frac{1}{60}$.$\frac{\frac{1}{a+1}}{a + \frac{1}{2}} = \frac{1}{60}$.$\frac{1}{(a+1)(\frac{2a+1}{2})} = \frac{1}{60} \Rightarrow (a+1)(2a+1) = 30$.$2a^2 + 3a + 1 = 30 \Rightarrow 2a^2 + 3a - 29 = 0$.Wait,re-evaluating the denominator sum: $\sum_{r=1}^n (na+r) = n^2a + \frac{n(n+1)}{2}$. Dividing by $n^2$ gives $a + \frac{1}{2}(1 + \frac{1}{n}) 	o a + \frac{1}{2}$.Correcting the equation: $\frac{1/(a+1)}{a + 1/2} = 1/60$ $\Rightarrow \frac{2}{(a+1)(2a+1)} = 1/60$ $\Rightarrow (a+1)(2a+1) = 120$.$2a^2 + 3a + 1 = 120 \Rightarrow 2a^2 + 3a - 119 = 0$.$(2a + 17)(a - 7) = 0$.Since $a > 0$,we have $a = 7$.

If $\mathop {\lim }\limits_{n \to \infty } \frac{{{1^a} + {2^a} + \dots + {n^a}}}{{{{\left( {n + 1} \right)}^{a - 1}}\left[ {\left( {na + 1} \right) + \dots + \left( {na + n} \right)} \right]}} = \frac{1}{{60}}$ for some positive real number $a$,then $a$ is equal to

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