If $\lim _{n \rightarrow \infty} \frac{(n+1)^{k-1}}{n^{k+1}}[(n k+1)+(n k+2)+\ldots+ (n k+n)]=33 \cdot \lim _{n \rightarrow \infty} \frac{1}{n^{k+1}} \cdot\left[1^{k}+2^{k}+3^{k}+\ldots+n^{k}\right]$, then the integral value of k is equal to ______.

Limits — JEE Main · 2022 · 25 Jul 2022 · Shift 1 Question with Solution

lim_{n \to \infty} \frac{( n + 1 ) ^{k - 1}}{n ^{k + 1}} [(nk + 1) + (nk + 2) + \dots + (nk + n)] = 33 \cdot lim_{n \to \infty} \frac{1}{n ^{k + 1}} \cdot [1^{k} + 2^{k} + 3^{k} + \dots + n^{k}]

, then the integral value of k is equal to ______.

Answer: 5

LHS: The sum

(nk + 1) + (nk + 2) + \dots + (nk + n)

has

n

terms with average

nk + \frac{n + 1}{2}

, so the sum is

n^{2} k + \frac{n ( n + 1 )}{2}

n \to \infty lim \frac{( n + 1 ) ^{k - 1}}{n ^{k + 1}} \cdot [n^{2} k + \frac{n ( n + 1 )}{2}] = n \to \infty lim \frac{( n + 1 ) ^{k - 1} \cdot n ^{2}}{n ^{k + 1}} (k + \frac{1 + \frac{1}{n}}{2})

Simplifying:

\frac{( n + 1 ) ^{k - 1} \cdot n ^{2}}{n ^{k + 1}} = \frac{( n + 1 ) ^{k - 1}}{n ^{k - 1}} \to 1

n \to \infty

.

Thus LHS

= k + \frac{1}{2}

.

RHS: Using the standard result

\sum_{i = 1}^{n} i^{k} \sim \frac{n ^{k + 1}}{k + 1}

n \to \infty

n \to \infty lim \frac{1}{n ^{k + 1}} i = 1 \sum n i^{k} = \frac{1}{k + 1}

So RHS

= 33 \cdot \frac{1}{k + 1} = \frac{33}{k + 1}

.

Equating:

k + \frac{1}{2} = \frac{33}{k + 1}

(2 k + 1) (k + 1) = 66

2 k^{2} + 3 k + 1 = 66

2 k^{2} + 3 k - 65 = 0

(k - 5) (2 k + 13) = 0

Since

k

must be a positive integer,

k = 5

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Integer answer