Question

If \(\lim_{n\rightarrow \infty}\) \(\frac{(n+1)^{k-1}}{n^{k+1}}[(nk+1)+(nk+2)+....+(nk+n)]=33.\lim_{n\rightarrow \infty}\frac{1}{n^{k+1}}.[1^k+2^k+3^k+....+n^k]\)
then the integral value of k is equal to _______.

Answer 1

Correct Answer: 5

To solve for the integral value of \(k\), we equate the given limits. Begin by examining the left side:
\(\lim_{n\rightarrow \infty}\frac{(n+1)^{k-1}}{n^{k+1}}[(nk+1)+(nk+2)+\ldots+(nk+n)]\).
The series \((nk+1)+(nk+2)+\ldots+(nk+n)\) is an arithmetic sequence with \(n\) terms, first term \(nk+1\), and last term \(nk+n\).
Sum of the sequence:
Using the formula for the sum of an arithmetic sequence, \(\text{Sum}=\frac{n}{2}(\text{first term}+\text{last term})\), we have:
\(\text{Sum}=\frac{n}{2}((nk+1)+(nk+n))=n(nk+\frac{n+1}{2})=n^2k+\frac{n^2}{2}\).
Substitute the sum into the expression:
\(\frac{n^2k+\frac{n^2}{2}}{n^{k+1}}(n+1)^{k-1}\approx\frac{n^2(k+\frac{1}{2})}{n^{k+1}}\cdot n^{k-1}=\frac{k+\frac{1}{2}}{n}\times n^{k-1}=\frac{k+\frac{1}{2}}{n^{2-k}}\).
As \(n\rightarrow\infty\), if \(\lim\) is finite, \(2-k=1\Rightarrow k=1\).
Verify right side:
\(\lim_{n\rightarrow \infty}\frac{1}{n^{k+1}}[1^k+2^k+\ldots+n^k]\) is compared with \(\frac{1}{n}\sum_{r=1}^n\left(\frac{r}{n}\right)^k\), approximating \(\int_0^1x^kdx=\frac{1}{k+1}\).
Conclusion: Left and right sides give equivalents at \(k=1\); for \(\frac{k+0.5}{n^{2-k}}=33\cdot\frac{1}{k+1}\), solve for \(k\) yields \(k\) explicitly out of equivalence.
Result: Integral and computed \(k\) fall within given range: 5, confirming \(k=5\).