Question

$\text{If}\underset{n\to \infty }{lim}\frac{(1^{\lambda }+2^{\lambda }+3^{\lambda }+--+n^{\lambda })}{(1^4+2^4+--+n^4)(1^3+2^3+--+n^3)}=F\left(\lambda \right),\lambda ϵN,\text{then}$

• A. $F\left(\lambda \right)$ is finite for $\lambda \le 8$
• B. $F\left(8\right)=\frac{20}{11}$
• C. $F\left(8\right)=\frac{20}{9}$
• D. $F\left(7\right)=0$

Answer 1

Answer: (A), (C), (D)

$F\left(7\right)=0$

$F\left(\lambda \right)=\underset{n\to \infty }{lim}\frac{{\sum }_{l=1}^n{l}^{\lambda }}{\left({\sum }_{l=1}^n{l}^4\right)\left({\sum }_{l=1}^n{l}^3\right)}=\frac{(\frac{1}{n}\sum (\frac{l}{n}{)}^{\lambda }).n^{\lambda +1}}{\left(\frac{1}{n}{\sum }_{l=1}^n\right(\frac{l}{n}{)}^4\left)\right(\frac{1}{n}\left({\sum }_{l=1}^n\right(\frac{l}{n}{)}^3)n^9}=\underset{n\to \infty }{lim}\frac{{\int }_0^1{x}^{\lambda }dx}{\left({\int }_0^1{x}^{4}dx\right)\left({\int }_0^1{x}^{3}dx\right)}.n^{\lambda -8}\Rightarrow \frac{20}{\lambda +1}\times \frac{1}{n^{8-\lambda }}=F\left(\lambda \right)$