Question

For \( a \in \mathbb{R} \), \( a \neq -1 \), \( \lim_{n \to \infty} \frac{1^a + 2^a + \dots + n^a}{(n+1)^{a-1}[(na+1)+(na+2)+\dots+(na+n)]} = \frac{1}{60} \). Then one of the values of a is

• A. 5
• B. 8
• C. \( -\frac{15}{2} \)
• D. \( -\frac{17}{2} \)

Hint:

For limits involving sums as \( n \to \infty \), use integral approximations. The sum \( \sum_{k=1}^n k^p \) can be approximated by \( \int_0^n x^p dx = \frac{n^{p+1}}{p+1} \). For arithmetic series, approximate the sum by taking the number of terms times the average of the first and last terms, ignoring smaller constants for large n.

Answer 1

The Correct Option is D

Let's simplify the numerator and denominator separately. 

Numerator (N): \( \sum_{k=1}^n k^a \). Using the integral approximation for large n: \[ N \approx \int_0^n x^a dx = \left[\frac{x^{a+1}}{a+1}\right]_0^n = \frac{n^{a+1}}{a+1} \] 

Denominator (D): Let's simplify the two parts. 

Part 1: \( (n+1)^{a-1} \). For large n, this is approximately \( n^{a-1} \). 

Part 2: The sum \( S = (na+1)+(na+2)+\dots+(na+n) \). 

This is an arithmetic progression with n terms. The sum is \( S = \frac{n}{2}(\text{first term} + \text{last term}) = \frac{n}{2}(na+1 + na+n) \). For large n, the constants are negligible, so \( S \approx \frac{n}{2}(na+na+n) = \frac{n}{2}(2na+n) = \frac{n^2}{2}(2a+1) \). So, the denominator is approximately \( D \approx n^{a-1} \cdot \frac{n^2}{2}(2a+1) = \frac{n^{a+1}}{2}(2a+1) \). Now let's evaluate the limit of the fraction \( \frac{N}{D} \): \[ \lim_{n \to \infty} \frac{\frac{n^{a+1}}{a+1}}{\frac{n^{a+1}}{2}(2a+1)} = \frac{1/(a+1)}{(2a+1)/2} = \frac{2}{(a+1)(2a+1)} \] We are given that this limit is equal to \( \frac{1}{60} \). \[ \frac{2}{(a+1)(2a+1)} = \frac{1}{60} \] \[ (a+1)(2a+1) = 120 \] \[ 2a^2 + 3a + 1 = 120 \] \[ 2a^2 + 3a - 119 = 0 \] We can solve this quadratic equation. 

Let's test the options. 

(A) a=5: \( 2(25)+3(5)-119 = 50+15-119 \neq 0 \) 

(B) a=8: \( 2(64)+3(8)-119 = 128+24-119 \neq 0 \)

(C) a=-15/2: \( 2(225/4)+3(-15/2)-119 = 225/2 - 45/2 - 238/2 = (180-238)/2 \neq 0 \) 

(D) a=-17/2: \( 2((-17/2)^2)+3(-17/2)-119 = 2(289/4) - 51/2 - 119 = 289/2 - 51/2 - 238/2 = (238-238)/2 = 0 \). This value satisfies the equation.