Limit of Sequence of Powers and Linear Sums

To find the value of $a$, let the given limit be $L$: $L = \lim_{n \to \infty} \frac{1^a + 2^a + \dots + n^a}{(n+1)^{a-1} [(na+1) + (na+2) + \dots + (na+n)]} = \frac{1}{60}$

Step 1: Simplify the Denominator

The sum in the square brackets forms an arithmetic progression with $n$ terms: $\sum_{r=1}^n (na + r) = n(na) + \sum_{r=1}^n r = n^2 a + \frac{n(n+1)}{2} = \frac{n}{2} [2na + n + 1] = \frac{n^2}{2} \left(2a + 1 + \frac{1}{n}\right)$

Thus, the denominator $D_n$ can be expressed as: $D_n = (n+1)^{a-1} \cdot \frac{n^2}{2} \left(2a + 1 + \frac{1}{n}\right) = \frac{n^{a+1}}{2} \left(1 + \frac{1}{n}\right)^{a-1} \left(2a + 1 + \frac{1}{n}\right)$

As $n \to \infty$: $D_n \sim \frac{2a+1}{2} n^{a+1}$

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Step 2: Evaluate the Numerator and the Limit

Case 1: $a > -1$ When $a > -1$, we can express the numerator $S_n = \sum_{r=1}^n r^a$ as a Riemann sum: $\lim_{n \to \infty} \frac{1}{n^{a+1}} \sum_{r=1}^n r^a = \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^n \left(\frac{r}{n}\right)^a = \int_0^1 x^a \, dx = \left[ \frac{x^{a+1}}{a+1} \right]_0^1 = \frac{1}{a+1}$

Now, substituting the behavior of both numerator and denominator into the limit: $L = \lim_{n \to \infty} \frac{\frac{1}{n^{a+1}} \sum_{r=1}^n r^a}{\frac{1}{2} \left(1 + \frac{1}{n}\right)^{a-1} \left(2a + 1 + \frac{1}{n}\right)} = \frac{\frac{1}{a+1}}{\frac{2a+1}{2}} = \frac{2}{(a+1)(2a+1)}$

Equating this to the given value $\frac{1}{60}$: $\frac{2}{(a+1)(2a+1)} = \frac{1}{60}$ $(a+1)(2a+1) = 120$ $2a^2 + 3a + 1 = 120$ $2a^2 + 3a - 119 = 0$

Solving this quadratic equation for $a$: $a = \frac{-3 \pm \sqrt{3^2 - 4(2)(-119)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 952}}{4} = \frac{-3 \pm \sqrt{961}}{4} = \frac{-3 \pm 31}{4}$

This yields two roots: $a = \frac{28}{4} = 7 \quad \text{or} \quad a = -\frac{34}{4} = -\frac{17}{2}$

Since this case requires $a > -1$, the valid solution here is: $a = 7$

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Case 2: $a < -1$ If $a = -\frac{17}{2} < -1$, the series in the numerator converges to a finite positive constant by the $p$-series test: $\lim_{n \to \infty} \sum_{r=1}^n r^{-17/2} = \zeta\left(\frac{17}{2}\right) \in (1, \infty)$ However, the denominator behaves as: $D_n \sim \frac{2(-17/2)+1}{2} n^{-17/2 + 1} = -8 n^{-15/2} \to 0 \quad \text{as } n \to \infty$

Consequently, for $a = -\frac{17}{2}$: $L = \lim_{n \to \infty} \frac{\sum_{r=1}^n r^{-17/2}}{D_n} = \pm \infty \neq \frac{1}{60}$ Thus, $a = -\frac{17}{2}$ is an extraneous solution.

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Conclusion

The only valid value of $a$ is $7$.

Therefore, the correct option is B.