Question

${lim}_{n\to \infty }\frac{(1^k+2^k+3^k+.....+n^k)}{(1^2+2^2+.....+n^2)(1^3+2^3+.....+n^3)}=F\left(k\right)$, then $(k\in N)$

• A. $F\left(k\right)isfinitefork\le 6$
• B. $F\left(5\right)=0$
• C. $F\left(6\right)=\frac{12}{7}$
• D. $F\left(6\right)=\frac{5}{7}$

Answer 1

The correct option is A $F\left(k\right)isfinitefork\le 6$

$1^2+2^2+...+n^2=\left(\frac{n(n+1)(2n+1)}{6}\right)=max.power{n}^{3}andincaseof{1}^3+2^3+......+n^3={\left[\right(\frac{n(n+1)}{2}\left)\right]}^2=max.power{n}^4$

so the maximum power is one more than the power raised to natural numbers in the sequence and total max. power raised in denominator is 7

so for value of K $⩽$ 6, maximum power raised in the numerator will be $⩽$ 7 and then limit will have a finite value.