# limit of a series

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

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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The numerator and the denominator can be expressed by a closed form. For example : $1^3 + 4^3 + ... + (3n-2)^3 = \sum_{k=1}^{n} (3k-2)^3 = 27 \sum k^3 -54 \sum k^2 + 36 \sum k + 8n$... –  Paul Pichaureau Feb 5 '12 at 15:29
About the denominator: you don't "sum to infinity" and then multiply by itself. Ideally, you do this operation normally, multiplying that finite quantity by itself (for a generic $n$), and then compute the limit as $n \rightarrow \infty$. It may be the case, however, that some simplification can be done first. –  student Feb 5 '12 at 15:30
yeah i knew that for the top part. but can I just do the same thing for the denominator? Isn't it different because the power 2 is outside of the summation? –  nofe Feb 5 '12 at 15:31
@nofe: no. The exponent $2$ is outside the summation, I agree, but it's not outside the limit, so you should in principle compute it before taking the limit. By the way, there's no such things as summing infinitely many numbers. What we mean by that is always really a limit. –  student Feb 5 '12 at 15:34
What makes you hesitate? You can just sum and square it. Indeed, direct summing gives $$1 + 4 + 7 + \cdots + (3n-2) = \frac{3n^2 -n}{2},$$ so the denomiator is $\frac{1}{4} n^2 (3n-1)^2$. –  sos440 Feb 5 '12 at 15:38
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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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i didn't know there were these formulas for summation. Can I derive them or is it something to memorize? –  nofe Feb 5 '12 at 16:48
@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. –  Henning Makholm Feb 5 '12 at 16:56

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 $3$.

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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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There is an error : $(3k-2)^3 = 3^3k^3 - 3*3^2*2*k^2+3*3*2^2*k - 2^3 = 27k^3-54*k+36*k-8$ –  Paul Pichaureau Feb 5 '12 at 16:05
@Paul You're absolutely right. My head missed it! I am sorry. I'll fix it now! –  user21436 Feb 5 '12 at 16:11
Just lazy to correct other steps, but I think I haven't made it all cryptic. @PaulPichaureau –  user21436 Feb 5 '12 at 16:14
A second little mistake is in the denominator ($3^2$ instead of $27=3^3$ I think...). Sorry. –  Raymond Manzoni Feb 5 '12 at 16:50
@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! –  user21436 Feb 5 '12 at 17:10