$\large \lim_{n\to \infty} \frac{1^6+2^6+\cdots+n^6}{(1^2+2^2+\cdots+n^2)(1^3+2^3+\cdots+n^3)}$

If the above limit can be expressed as $\dfrac ab$ , where $a$ and $b$ are coprime integers, find $a+b$ .

The answer is 19.

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There should be （1＋1/n）∧（u＋1），is that right？

yi cx
- 2 years, 2 months ago

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

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

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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 $12 + 7 = \boxed{19}$ .

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Is this what you were looking for @Vilakshan Gupta

James Harbour
- 2 years, 1 month ago

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

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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.

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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 $\frac{12}{7}$

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Do you remember the formula or derived it..? If derived then please teach me too because I can't understand the theory of this...(whatever faulhaber formula is) with... But this question was not meant to be solved that way... It can be calculated by simple integration ... But I want to know the exact correct solution by integration because you can't state it randomly...like integrate x^6 dx / integrate x^2 dx times integral x^3 dx will give you the answer.... what is the exact logic behind it...

Vilakshan Gupta
- 2 years, 2 months ago

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Well, I remembered the formula......You see, it always helps to remember such common things.....!! And as for the derivation, open class 11th Math NCERT and see the derivation of sum of squares formula in the chapter Sequences and Series.......You can easily proceed on with that method....!!

Regarding the integration method, yes I agree it is simple!!! Here is the method.....

Dividing the numerator and denominator by $n^7$ and using the rule that limit of a fraction is limit of the numerator divided by limit of the denominator, we get the expression in the numerator as

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

and the denominator as the product of following two expressions

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

and

$\displaystyle \frac{1}{n}\sum_{k=1}^{\infty}\left(\frac{k}{n}\right)^3$

Observe that when we take the limit as $n$ tends to $\infty$ the expressions in the fractions are simply the Riemann Sums of the functions $x^6$ , $x^2$ and $x^3$ respectively over $\left(0,1\right)$

So, the limit simply becomes

$\displaystyle \frac{\int_0^1x^6dx}{\left(\int_0^1x^3dx\right)\left(\int_0^1x^2dx\right)}=\frac{12}{7}$

Aaghaz Mahajan
- 2 years, 2 months ago

I know the method of calculating it by telescoping series... but it's lengthy...so wanted to know other method......Thanks for explaining integration method bro..... but seriously do you remember summation of 6th power of numbers..... It's difficult also and also not needed anywhere... have u learned all power summations......they get too complex ahead.....

Vilakshan Gupta
- 2 years, 2 months ago

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Well.........hehe......I don't think learning them is difficult!! Mujhe to mazza aata hai bro.......and obvo I HAVE NOT learned all the power summations!! There are infinite of them!! XD But, well, Faulhauber to mujhe power 7 tak ka yaad hai........TELESCOPING SERIES???!!! Seriously??!!! VOh to hadddd se zyaada lamba nhi ho jaayega kya??? Chal glad that you understood my explanation......NCERT kholi???

Aaghaz Mahajan
- 2 years, 2 months ago

Bhai telescopic aur ncert ek hi to hai.... Usme telescopic se hi to bataya hai....

Vilakshan Gupta
- 2 years, 2 months ago

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Ohh lol!! Tu voh keh rhaa tha..........mujhe laga you meant breaking down into partial fractions and some other stuff!!! XD

Aaghaz Mahajan
- 2 years, 2 months ago

telescopic btw yehi hota hai na ki terms cancel hoti hain.... ya fir partial fraction hona zaroori hota hai for it to be called telescopic....

Vilakshan Gupta
- 2 years, 2 months ago

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Umm.....dekh, telescopic ke liye tujhe bas yeh dekhna hai that successive terms cancel off.....ab voh partial fractions hone pe dependent nhi hai......kisi aur tareeke se bhi cancel ho sakti hain....Jaise, consider sum of Sines in AP, ab voh bhi telescopic maani jaayegi, even thought there are no "partial fractions"....Hope it helps :)

Aaghaz Mahajan
- 2 years, 2 months ago

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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 $\boxed{19}$ .