Question

If $\displaystyle \lim _{n \rightarrow \infty} \frac{(n+1)^{k-1}}{n^{k+1}}[(n k+1)+(n k+2)+\ldots+ $
$ (n k+n)]=33 . $
$\displaystyle \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 _______

Answer 1

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