<s>[INST] $\lim_{n\to \infty} F(k)=\frac{(1^{k}+2^{k}+3^{k}+.....+n^{k})}{(1^{2}+2^{2}+3^{2}+.....+n^{2})(1^{3}+2^{3}+3^{3}+.....+n^{3})}$ Find F(5) and F(6) Find the value of F(5) & F(6).It is given that
$$F(k)= \lim_{n\to \infty} \frac{(1^{k}+2^{k}+3^{k}+.....+n^{k})}{(1^{2}+2^{2}+3^{2}+.....+n^{2})(1^{3}+2^{3}+3^{3}+.....+n^{3})}$$ where 'k' is a Natural number
My approach Denominator is $\frac{(n)^{3}*(n+1)^{3}*(2n+1)}{24}$ 
I dont know how to proceed from here as the Numerator has power in k. I also tried definite integral formula but not able to solve it
 [/INST] note that $$\sum_{i=1}^ni^2=\frac{1}{6}n(n+1)(2n+1)$$
$$\sum_{i=1}^n i^3=\frac{1}{4}n^2(n+1)^2$$
$$\sum_{i=1}^ni^5=\frac{1}{12}n^2(n+1)^2(2n^2+2n-1)$$
and we get $$\lim_{n\to \infty}\frac{\frac{1}{12}n^2(n+1)^2(2n^2+2n-1)}{\frac{1}{6}n(n+1)(2n+1)\frac{1}{4}n^2(n+1)^2}$$
the searched limit is $0$
note for the next:
$$\sum_{n=1}^n i^6=1/42\,n \left( n+1 \right)  \left( 2\,n+1 \right)  \left( 3\,{n}^{4}+6
\,{n}^{3}-3\,n+1 \right) 
$$
</s>