Question

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

Answer 1

Answer: 5

LHS
$\underset{n\to \infty }{\lim }\frac{(n+1{)}^{k-1}}{n^{k+1}}[nk\cdot n+1+2+\dots +n]$
$=\underset{n\to \infty }{\lim }\frac{(n+1{)}^{k-1}}{n^{k+1}}\cdot \left[n^{2}k+\frac{n(n+1)}{2}\right]$
$=\underset{n\to \infty }{\lim }\frac{(n+1{)}^{k-1}\cdot n^2\left(k+\frac{\left(1+\frac{1}{n}\right)}{2}\right)}{n^{k+1}}$
$\Rightarrow \underset{n\to \infty }{\lim }\left(1+\frac{1}{n}\right)\left(k+\frac{\left(1+\frac{1}{n}\right)}{2}\right)$
$\Rightarrow \left(k+\frac{1}{2}\right)$
$\text{ RHS }$
$\Rightarrow \underset{n\to \infty }{\lim }\frac{1}{n^{k+1}}\left(1^k+2^k+\dots +n^k\right)=\frac{1}{k+1}$
$\text{ LHS }=\text{ RHS }$
$\Rightarrow k+\frac{1}{2}=33\cdot \frac{1}{k+1}$
$\Rightarrow (2k+1)(k+1)=66$
$\Rightarrow (k-5)(2k+13)=0$
$\Rightarrow k=5\text{ or }-\frac{13}{2}$