$\lim_{n\to \infty} F(k)=\frac{(1^{k}+2^{k}+3^{k}+.....+n^{k})}{(1^{2}+2^{2}+3^{2}+.....+n^{2})(1^{3}+2^{3}+3^{3}+.....+n^{3})}$ 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
 A: 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, $\frac{12}{7} = \lim \limits_{ n \to \infty} \frac{24 n^{6-2}}{(6+1) (n+1)^3 (2 n+1)} \leq F(6) \leq \lim \limits_{n \to \infty } \frac{24 \left((n+1)^{6+1}-1\right)}{(6+1) n^3 (n+1)^3 (2 n+1)} = \frac{12}{7}$
A: 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:
$$\sum_{n=1}^n i^6=1/42\,n \left( n+1 \right)  \left( 2\,n+1 \right)  \left( 3\,{n}^{4}+6
\,{n}^{3}-3\,n+1 \right) 
$$
A: If you know about generalized harmonic numbers
$$\sum_{i=1}^n i^k=H_n^{(-k)}$$ for which the asymptotics is 
$$H_n^{(-k)}=n^k
   \left(\frac{n}{k+1}+\frac{1}{2}+O\left(\frac{1}{n^2}\right)\right)+
   \zeta (-k)\sim n^k
   \left(\frac{n}{k+1}+\frac{1}{2}\right)$$ which is the beginning of the form submitted by Jacob Bernoulli and published in $1713$ (see here).
Using the denominator you wrote, this makes 
$$F(k) \sim \frac{24 n^{k-3} \left(\frac{n}{k+1}+\frac{1}{2}\right)}{(n+1)^3 (2 n+1)}\sim \frac{12 }{1+k} n^{k-6}$$
Edit
You can make this simpler if you notice that, in Faulhaber formulae, for
$$S_k=\sum_{i=1}^n i^k $$ the largest power is $k+1$ and the coefficient is just $\frac 1 {k+1}$. So, for infinitely large values of $n$ $$S_k \sim \frac {n^{k+1}}{k+1}$$
$$\Phi=\frac{(1^{a}+2^{a}+\cdots +n^{a})}{(1^{b}+2^{b}+\cdots+n^{b})(1^{c}+2^{c}+\cdots+n^{c})}\sim \frac{ (b+1)(c+1)}{a+1}n^{a-(b+c+1)}$$
So


*

*$a<b+c+1 \implies\lim_{n\to \infty} \Phi=0$

*$a=b+c+1\implies\lim_{n\to \infty} \Phi=\frac{ (b+1)(c+1)}{a+1}$

*$a>b+c+1\implies\lim_{n\to \infty} \Phi= \infty$

A: Cesaro-Stolz is your friend here. Using this powerful theorem we have $$\lim_{n\to\infty} \frac{1}{n^{k+1}}\sum_{i=1}^{n}i^{k}=\lim_{n\to\infty}\frac{n^{k}}{n^{k+1}-(n-1)^{k+1}}=\frac{1}{k+1},k>0\tag{1}$$ For your question you need to divide the numerator and denominator by $n^{7}$. The denominator then tends to $(1/3)(1/4)=1/12$ and numerator tends to $0$ or $1/7$ depending on whether $k$ is $5$ or $6$. Thus $F(5)=0,F(6)=12/7$.
