For $a \in \mathbb{R}$ (the set of all real numbers), $a \neq-1$, $\lim _{\mathrm{n} \rightarrow \infty} \frac{\left(1^{a}+2^{a}+\ldots+\mathrm{n}^{a}\right)}{(n+1)^{a-1}[(n a+1)+(n a+2)+\ldots+(n a+n)]}=\frac{1}{60}$ Then $a=$

JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Limits XII

For

a \in R

(the set of all real numbers),

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

a =

Answer: (B)

Simplify the limit:

lim_{n \to \infty} \frac{1 ^{a} + 2 ^{a} + \dots + n ^{a}}{( n + 1 ) ^{a - 1} \cdot n ^{2} [ a + \frac{1 + 1/ n}{2} ]}

= lim_{n \to \infty} \frac{n ^{a} \cdot \frac{1}{n} \sum _{k = 1}^{n} ( k / n ) ^{a}}{n ^{a - 1} ( 1 + 1/ n ) ^{a - 1} \cdot n ^{2} [ a + \frac{1}{2} ]}

= lim_{n \to \infty} \frac{1}{( 1 + 1/ n ) ^{a - 1} [ a + \frac{1}{2} ]} \cdot \int_{0}^{1} x^{a} d x

= \frac{1}{a + \frac{1}{2}} \cdot \frac{1}{a + 1} = \frac{2}{( 2 a + 1 ) ( a + 1 )} = \frac{1}{60}

(2 a + 1) (a + 1) = 120

2 a^{2} + 3 a + 1 = 120

2 a^{2} + 3 a - 119 = 0

(a - 7) (2 a + 17) = 0

This gives

a = 7

a = - 17/2

.

However, for

a < - 1

, the numerator

\sum_{k = 1}^{n} k^{a}

diverges, making the limit undefined. Therefore, only

a = 7

is valid.

JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · One or more correct options