Question

The value of $\underset{n\to \infty }{lim}\frac{\sum _{i=1}^n(i{)}^{p+q+1}}{\sum _{i=1}^n(i{)}^q\sum _{i=1}^n(i{)}^p}$ is equal to (where $p,q\ge 1$)

• A. $\frac{pq}{p+q+2}$
• B. $\frac{(p+1)(q+1)}{p+q+2}$
• C. $\frac{(p-1)(q-1)}{p+q+1}$
• D. $\frac{(pq{)}^2}{p+q+1}$

Answer 1

The correct option is B $\frac{(p+1)(q+1)}{p+q+2}$

$\underset{n\to \infty }{lim}\frac{n^{p+q+1}\sum _{i=1}^n{\left(\frac{i}{n}\right)}^{p+q+1}}{n^p\cdot n^q\sum _{i=1}^n{\left(\frac{i}{n}\right)}^q\cdot \sum _{i=1}^n{\left(\frac{i}{n}\right)}^p}=\underset{n\to \infty }{lim}\frac{\frac{1}{n}\sum _{i=1}^n{\left(\frac{i}{n}\right)}^{p+q+1}}{\frac{i}{n}\sum _{i=1}^n{\left(\frac{i}{n}\right)}^q\cdot \frac{1}{n}\sum _{i=1}^n{\left(\frac{i}{n}\right)}^p}=\frac{\underset{n\to \infty }{lim}\frac{1}{n}\sum _{i=1}^n{\left(\frac{i}{n}\right)}^{p+q+1}}{\underset{n\to \infty }{lim}\frac{1}{n}\sum _{i=1}^n{\left(\frac{i}{n}\right)}^q\cdot \underset{n\to \infty }{lim}\frac{1}{n}\sum _{i=1}^n{\left(\frac{i}{n}\right)}^p}=\frac{\underset{0}{\overset{1}{\int }}{x}^{p+q+1}dx}{\underset{0}{\overset{1}{\int }}{x}^{q}dx\times \underset{0}{\overset{1}{\int }}{x}^{p}dx}=\frac{\frac{1}{p+q+2}}{\frac{1}{p+1}\cdot \frac{1}{q+1}}=\frac{(p+1)(q+1)}{p+q+2}$