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$\lim_{n→∞} \frac{1^3 +4^3 +7^3 + ... + (3n-2)^3}{[(1+4+7+...+(3n-2)]^2}$

I'm mostly confused because I'm not sure what the raising to the second power in the denominator means. Does this mean that after I sum all of the terms of (3n-2) from n to infinity I then multiply that by itself?

Do I need to use Riemann sums here?

I tried to do it with the squeeze theorem but ended up getting infinity on the right side each time.

(I've looked around in "similar questions" to see if this has been asked before and as far as I've seen I couldn't find anything that helps me with this in previously asked questions.)

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    You may find the answer with few computations : the numerator is equivalent to $\int (3n)^3 dn\ $ and the denominator to $\left(\int (3n) dn\right)^2$.2012-02-05

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There's nowhere in the expression you're being asked to "sum all of the terms of $(3n-2)$ from $n$ to infinity".

What the expression does say is for each particular $n$ to:

  1. Add up the terms from $1$ to $3n-2$ -- there are exactly $n$ of these -- this is a finite sum.
  2. Multiply the result with itself.
  3. Add up the cubes of the same finitely many numbers $1$ through $3n-2$.
  4. Divide the result form (3) by the result from (2).

This gives one result for each $n$, and then you're asked to find the limit of those results as $n$ goes toward infinity.

In order to evaluate that limit without doing infinitely much work, a good way to start would be to try to simplify the fraction insider the limit to something simpler. While simplifying, you can forget everything about $n$ going to infinity -- you're just looking for a nicer way to compute the value of the fraction for some partiucular, finite $n$.

Do you know how to sum finite arithmetic sequences such as the one inside $[...]^2$ -- that is, a generalization of the familiar $1+2+\cdots+n=\frac{n^2+n}2$? Do you know a similar formula for cubes of arithmetic sequences for use in the numerator?

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    @nofe: The trick in the case of a simple arithmetic progression is to change the order of the terms to pair them off two by two $1+4+\cdots+(3n-5)+(3n-2)=(1+3n-2)+(4+3n-5)+\cdots = (3n-1)+(3n-1)+\cdots = \frac{n}{2}(3n-1)$ For higher powers I usually just look up the formula, though in a pinch I could figure out what it is by remembering that it is always a polynomial in $n$ of degree one higher than the power in the terms.2012-02-05
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Let's first compute what is inside the limit:

$\begin{align*}\dfrac{\sum_{k=1}^n(3k-2)^3}{\left(\sum_{k=1}^n 3k-2\right)^2}&=\dfrac{\sum27k^3-54k^2+36k-8}{\left(3\dfrac{n(n+1)}{2}-2n\right)^2} \\&=\dfrac{27\left(\dfrac{n(n+1)}{2}\right)^2-54\dfrac{n(n+1)(2n+1)}{6}+36\dfrac{n(n+1)}{2}-8n}{\dfrac{n^2}{4}\cdot(3n-1)^2}\\&=\dfrac{27n^4+\Theta(n^3)}{9n^4+\Theta(n^3)}\end{align*}$

So, the required limit is $3$.

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    @Raymond, why sorry? You're absolutely right. I am feeling upset now. I am typing out a set of notes. I don't know how many errors have crept in there already! Thank You!2012-02-05
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Fast computation :
the numerator is equivalent to $\int (3n)^3 dn=\frac{3^3}4 n^4$
the denominator is equivalent to $(\int 3n dn)^2=3^2(\frac12 n^2)^2=\frac{3^2}4 n^4$

so that the limit is $3$.