$\lim_{n\to \infty} F(k)=\frac{(1^{k}+2^{k}+3^{k}+.....+n^{k})}{(1^{2}+2^{2}+3^{2}+.....+n^{2})(1^{3}+2^{3}+3^{3}+.....+n^{3})}$ 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

For $f$ increasing function we have that $ \int \limits_{a-1}^{b} f(s)ds \leq \sum \limits_{s=a}^{b} f(s) < \int \limits_{a}^{b+1} f(s) ds$
So $ \lim \limits_{n \to \infty} \frac{\int \limits_{0}^{n} s^k ds}{\sum \limits_{s=1}^{n} s^2 \sum \limits_{s=1}^{n} s^3}< F(k) <\lim \limits_{n \to \infty} \frac{\int \limits_{1}^{n+1} s^k ds}{\sum \limits_{s=1}^{n} s^2 \sum \limits_{s=1}^{n} s^3}$
Which is $ \lim \limits_{ n \to \infty} \frac{24 n^{k-2}}{(k+1) (n+1)^3 (2 n+1)} \leq F(k) \leq \lim \limits_{n \to \infty } \frac{24 \left((n+1)^{k+1}-1\right)}{(k+1) n^3 (n+1)^3 (2 n+1)}$
So,  $0= \lim \limits_{ n \to \infty} \frac{24 n^{5-2}}{(5+1) (n+1)^3 (2 n+1)}\leq F(5) \leq \lim \limits_{n \to \infty } \frac{24 \left((n+1)^{5+1}-1\right)}{(5+1) n^3 (n+1)^3 (2 n+1)} = 0$
And, $\frac{12}{7} = \lim \limits_{ n \to \infty} \frac{24 n^{6-2}}{(6+1) (n+1)^3 (2 n+1)} \leq F(6) \leq \lim \limits_{n \to \infty } \frac{24 \left((n+1)^{6+1}-1\right)}{(6+1) n^3 (n+1)^3 (2 n+1)} = \frac{12}{7}$
