6th power summation

Since $x \mapsto x^u$ is an increasing function of $x \ge 0$ for all $u > 0$ , we deduce that $\int_{r-1}^r x^u\,du \; \le \; r^u \; \le \; \int_r^{r+1}x^u\,du$ for any integer $r \ge 1$ , and hence $\frac{n^{u+1}}{u+1} \; = \; \int_0^n x^u\,du \; \le \; \sum_{r=1}^n r^u \; \le \; \int_1^{n+1} x^u\,du \; = \; \frac{(n+1)^{u+1}-1}{u+1} \; < \; \frac{(n+1)^{u+1}}{u+1}$ for all integers $n \ge 1$ , so that $\frac{1}{u+1} \; \le \; \frac{1}{n^{u+1}}\sum_{r=1}^n r^u \; \le \; \frac{1}{u+1}\left(1 + \tfrac{1}{n}\right)^{u+1}$ for all integers $n\ge 1$ , and hence $\lim_{n \to \infty}\frac{1}{n^{u+1}}\sum_{r=1}^n r^u \; =\; \frac{1}{u+1}$ Thus we have shown that $S_u(n) \; = \; \sum_{r=1}^n r^u \; \sim \; \tfrac{1}{u+1}n^{u+1} \hspace{2cm} n \to \infty$ and hence $\frac{S_{u+v+1}(n)}{S_u(n)S_v(n)} \; \sim \; \frac{\frac{1}{u+v+2}n^{u+v+2}}{\frac{1}{u+1}n^{u+1} \times \frac{1}{v+1}n^{v+1}} \; = \; \frac{(u+1)(v+1)}{u+v+2} \hspace{2cm} n \to \infty$ and hence $\lim_{n \to \infty} \frac{S_{u+v+1}(n)}{S_u(n)S_v(n)} \; = \; \frac{(u+1)(v+1)}{u+v+2}$ for any $u,v > 0$ . In this case we have $u=2$ , $v=3$ , so that the limit is $\tfrac{12}{7}$ , making the answer $\boxed{19}$ .

Here is a solution relying on a bit of mathematical intuition and trusting that the 'tetchy bits and bobs ' will look after themselves. As n tends to infinity, each of the sums are more and more closely approximated by the area under the graphs $y=x^n$ for n=6,2 or 1. Carrying out the integrations gives

$\frac{a}{b}=\lim_{n\to\infty}\frac{\frac{n^7}{7}}{\frac{n^3}{3}\frac{n^4}{4}}=\frac{12}{7}$

This problem, after some simplification, boils down to interpreting the limit of a Riemann Sum as a definite integral.

To begin, we simplify the denominator with the common formulas for their summations: $\lim_{n\to\infty}\frac{\displaystyle\sum_{k=1}^{n}k^6}{\left(\frac{n(n+1)(2n+1)}{6}\right)\cdot\left(\frac{n(n+1)}{2}\right)^2}$

We then perform a bit of algebraic manipulation to take advantage of our limit: $24 \cdot \lim_{n \to \infty} \displaystyle\sum_{k=1}^{n} \frac{k^6}{n^3(n + 1)^3(2n + 1)} \\ 24 \cdot\lim_{n\to\infty}\frac{1}{n}\displaystyle\sum_{k=1}^{n}\dfrac{k^6}{n^2 \left[n^3\left(1+\dfrac{3}{n}+\dfrac{3}{n^2}+\dfrac{1}{n^3}\right)\right]\left[2n\left(1+\dfrac{1}{2n}\right)\right]}$

Because the limit approaches infinity, all of the terms with a power of $n$ in the denominator will approach zero, so we can "ignore" them: $24 \cdot \lim_{n \to \infty} \displaystyle\sum_{k=1}^{n} \frac{k^6}{n^2 \cdot n^3 \cdot 2n} = 24 \cdot \lim_{n \to \infty} \displaystyle\sum_{k=1}^{n} \frac{k^6}{2n^6} \\$

Now we have a limit of a Riemann Sum, which we can interpret as a definite integral by letting $x=\dfrac{k}{n}$ and taking the result as our integrand with some bounds $a$ and $b$ : $24 \cdot \lim_{n \to \infty} \displaystyle\sum_{k=1}^{n} \frac{k^6}{2n^6} = 24 \cdot \int_a^b \dfrac{x^6}{2} \textrm{d}x$

To calculate the bounds, we take the limit of our substitution using the summation's first and last indices as $k$ : $a=\lim_{n\to\infty}\dfrac{1}{n}=0 \\ b=\lim_{n\to\infty}\dfrac{n}{n}=1$

Now we substitute these values and solve the resulting integral: $24 \cdot \int_0^1 \dfrac{x^6}{2} \textrm{d}x = \dfrac{12}{7}$

And thus, the answer is $12 + 7 = \boxed{19}$ .

Let $s_{a,n} = \displaystyle \sum_{k=1}^n k^a$ . Then the limit can be rewritten as $\displaystyle L = \lim_{n \to \infty} \frac {s_{6,n}}{s_{2,n}s_{3,n}}$ . It is commonly known that $s_{2,n} = \displaystyle \sum_{k=1}^n k^2 = \frac {n(n+1)(2n+1)}6$ and $s_{3,n} = \displaystyle \sum_{k=1}^n k^3 = \frac {n^2(n+1)^2}4$ .

For $s_{6,n}$ , we can use the Faulhaber's formula $\displaystyle \sum_{k=1}^n k^a = \frac1{a+1} \sum_{j=0}^{a} (-1)^j \binom{a+1}{j} B_j n^{a+1-j}$ , where $B_n$ denotes the $n$ th Bernoulli number as follows:

$\begin{aligned} \sum_{k=1}^n k^6 & = \frac 17 \sum_{j=0}^6 (-1)^j \binom 7j B_j n^{7-j} \\ & = \frac 17 \left(n^7 + \frac 72n^6 + \frac 72n^5 - \frac 76n^3 + \frac 16n \right) \\ & = \frac 1{42} \left(6n^7 + 21n^6 + 21n^5 - 7n^3 + n \right) \\ & = \frac {n \left(6n^6 + 21n^5 + 21n^4 - 7n^2 + 1 \right)}{42} \\ & = \frac {n(n+1)\left(6n^5 + 15n^4 + 6n^3 - 6n^2 - n + 1 \right)}{42} \\ & = \frac {n(n+1)(2n+1)\left(3n^4 + 6n^3 - 3n + 1 \right)}{42} \end{aligned}$

Then we have:

$\begin{aligned} L & = \lim_{n \to \infty} \frac {\frac 1{42}n(n+1)(2n+1)\left(3n^4 + 6n^3 - 3n + 1\right)}{\frac 16n(n+1)(2n+1) \times \frac 14n^2(n+1)^2} \\ & = \lim_{n \to \infty} \frac {4\left(3n^4 + 6n^3 - 3n + 1\right)}{7n^2(n+1)^2} \\ & = \lim_{n \to \infty} \frac {4\left(3n^4 + 6n^3 - 3n + 1\right)}{7\left(n^4+2n^3+n^2\right)} & \small \color{#3D99F6} \text{Divide up and down by }x^4 \\ & = \lim_{n \to \infty} \frac {4\left(3 + \frac 6n - \frac 3{n^3} + \frac 1{n^4} \right)}{7\left(1+ \frac 2n + \frac 1{n^2} \right)} \\ & = \frac {12}7 \end{aligned}$

Therefore, $a+b = 12+7 = \boxed{19}$ .

As noted in other solutions, the key is to show that the function $f_p(n)=\sum^n_{k=1} k^p$ is a polynomial in $n$ of degree $p+1$ with leading coefficient $\frac{1}{p+1}$ . Letting $n\rightarrow \infty$ in the given expression, we then find the limit is $\frac{\frac{1}{7}}{\frac{1}{3}\cdot\frac{1}{4}}=\frac{12}{7}$ , giving the answer $12+7=\boxed{19}$ .

A simple way of deriving the leading term of a polynomial is the method of differences ; the first differences of the function $f_p(k)$ are, by definition, $k^p$ . The $(p+1)^{th}$ differences are constant, at $p!$ . This tells us the function $f_p$ is indeed a polynomial of degree $p+1$ , and its leading coefficient is $\frac{p!}{(p+1)!}=\frac{1}{p+1}$ , as required.

This is a simple application of Faulhaber's formula .

We know that

$\displaystyle \sum_{k=1}^nk^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$

$\displaystyle \sum_{k=1}^nk^3=\frac{n^2\left(n+1\right)^2}{4}$

and

$\displaystyle \sum_{k=1}^nk^6=\frac{n\left(n+1\right)\left(2n+1\right)\left(3n^4+6n^3-3n+1\right)}{42}$

Putting these expressions in the limit and dividing the numerator and denominator by $n^4$ we get the answer to be $\frac{12}{7}$