Evaluate limit containing $\sum{n^6}$ Evaluate:
$$\lim_{n\to\infty}{\frac{1^6+2^6+3^6+\ldots+n^6}{(1^2+2^2+3^2+\ldots+n^2)(1^3+2^3+3^3+\ldots+n^3)}}$$
I can solve the denominator as:
$$\frac{n(n+1)(2n+1)}{6}\cdot\frac{n^2(n+1)^2}{4}$$
$$n^7\cdot\frac{(1+\frac{1}{n})(2+\frac{1}{n})}{6}\cdot\frac{(1+\frac{1}{n})}{4}$$
$$=\frac{n^7}{12}$$
How can I reduce the numerator?
 A: The sum of the $k^{th}$ power of the integers is a polynomial of degree $k+1$ in $n$, with leading term $\dfrac{n^{k+1}}{k+1}$.
Indeed, by the binomial theorem,
$$n^k=\sum_{i=1}^ni^k-\sum_{i=1}^{n-1}i^k\sim\frac{n^{k+1}}{k+1}-\frac{(n-1)^{k+1}}{k+1}=n^k+\text{lower degree terms}.$$
The given limit equals that of
$$\frac{n^7\cdot3\cdot4}{7\cdot n^3\cdot n^4}.$$
A: Note that
$$\lim_{n\to\infty}{\frac{1^6+2^6+3^6...+n^6}{(1^2+2^2+3^2...+n^2)(1^3+2^3+3^3...+n^3)}}=\lim_{n\to\infty} \frac{12\sum_1^n k^6}{n^7}=\frac{12}{7}$$
indeed by Stolz-Cesaro
$$\lim_{n\to\infty} \frac{12\sum_1^n k^6}{n^7}=\lim_{n\to\infty} \frac{12(n+1)^6}{(n+1)^7-n^7}=\lim_{n\to\infty} \frac{12(n+1)^6}{7n^6+...+1}=\frac{12}{7}$$
A: You wish to derive a formula for the sum $\sum_{k=1}^n k^6$ using "high-school methods".
Consider $$\frac{x^7}7-\frac{(x-1)^7}7=x^6-3x^5+5x^4+\cdots\\\frac{x^6}6-\frac{(x-1)^6}6=x^5-\cdots\\\cdots$$
So, defining $$I_n:=\frac{n^7}7+\alpha_1\frac{n^6}6+\alpha_2\frac{n^5}5+\cdots+\alpha_6 n+\alpha_7$$ for some constants $\alpha_i$, you can eliminate all the other factors other than $n^6$. So you end up with $$I_n-I_{n-1}=n^6\\\implies \sum_{k=1}^n k^6=I_n-I_0=\frac{n^7}7+\text{lower order terms}$$
This is all that matters to taking your limit, since all the lower order terms go to zero when divided by the $n^7$ you have on the denominator.
A: Here is yet another answer, which does not require any of the full, explicit formulae for sums of powers.
Consider the sum
$$
S(n, p) = \sum_{m = 1}^{n} m^p\, .
$$
I can rewrite this as
$$
S(n, p) = n^{p+1}\sum_{m = 1}^{n} \frac{1}{n}\;\left(\frac{m}{n}\right)^p\, .
$$
The sum in the above expression is a Riemann sum which approximates the integral
$$
\sum_{m = 1}^{n}\frac{1}{n}\;\left(\frac{m}{n}\right)^p
\approx \int_0^1 x^p\, dx \;=\; \frac{1}{p+1}\, .
$$
Thus, for large $n$
$$
S(n, p) \approx \frac{n^{p+1}}{p+1}\, .
$$
We can use this expression to replace all three sums in your expression:
$$
\lim_{n\rightarrow\infty} \frac{1^6 + \;...\; + n^6}{\left(1^2 +\;...\;+ n^2\right) \left(1^3 +\;...\;+ n^3\right)}
\;=\;
\lim_{n\rightarrow\infty} \frac{n^7 / 7}{\left(n^3 / 3\right) \left(n^4 / 4\right)}
\;=\; \frac{12}{7}
$$
A: Using Cesaro-Stolz (or limit of a Riemann sum) one can easily prove that $$\lim_{n\to\infty} \frac{1^p+2^p+\dots +n^p} {n^{p+1}}=\frac{1}{p+1}, p>-1$$ For the current question divide the numerator and denominator by $n^7=n^3\cdot n^4$ and then the limit is equal to $1/7/((1/3)(1/4))=12/7$.
