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I need to find a limit of this sequence $$\lim_{n\to\infty}\frac{1^2+2^2+...+n^2}{n^3-n+13}$$

I've tried to use the squeeze theorem, but I could not find two sequences with the same limit. From Wolfram Alpha I learned that the limit is $\frac{1}{3}$, but I just can't find a way to get a sequence greater or less than this one, that would have this limit.

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    $\begingroup$ Start by looking up a formula for $1^2+2^2+\dots+n^2$. Once you have it, the rest should be obvious. $\endgroup$ – Andrés E. Caicedo Nov 30 '13 at 22:24
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HINT:

$$\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6\;.$$

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    $\begingroup$ This is the best way to go, as it results in the simplest analysis. $\endgroup$ – ncmathsadist Nov 30 '13 at 22:36
  • $\begingroup$ I've solved it, thanks! I had no idea there is a formula for this. $\endgroup$ – Robert Nov 30 '13 at 22:36
  • $\begingroup$ @Robert: You’re welcome! If you’re taking a standard calculus course, you might find that formula in the section of your textbook that introduces Riemann sums. $\endgroup$ – Brian M. Scott Nov 30 '13 at 22:40
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L'Hopital it... For sequences the equivalent of L'Hopital rule is the Stolz Cezaro Theorem, which is this case says:

$$\lim_{n\to\infty}\frac{1^2+2^2+...+n^2}{n^3-n+13}=\lim_{n\to\infty}\frac{\left(1^2+2^2+...+n^2+(n+1)^2\right)-\left(1^2+2^2+...+n^2\right)}{\left((n+1)^3-(n+1)+13\right)-\left(n^3-n+13\right)}\\=\lim_{n\to\infty}\frac{(n+1)^2}{(n+1)^3-n^3+1}\\$$

Now both the denominator and numerator are quadratic polynomials.

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  • $\begingroup$ +1 because this method does not require you to know the sum. $\endgroup$ – MrYouMath Mar 6 '17 at 15:11
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Following Brian's solution, you expand for a cubic. Now think about this: As x reaches infinity, the terms don't really matter except for the x^3 terms. So the answer is...

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  • $\begingroup$ In my opinion this should have been a comment on B.'s answer. $\endgroup$ – Git Gud Nov 30 '13 at 22:41
  • $\begingroup$ Thanks for telling me. I couldn't comment until some person gave +1 to this answer, so I guess you can convert this to a comment. $\endgroup$ – pie314271 Nov 30 '13 at 22:44
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An alternative - more general - method uses the fact that $\sum_{k=1}^{n}k^p=\frac{1}{p+1}n^{p+1}+P(n),$ in which $P(n)$ is a polynomial of degree $p$. This can be shown by induction.

For this question $p=2$, hence $\frac{\sum_{k=1}^{n}k^2}{n^3-n+13}=\frac{\frac{1}{3}n^{3}+P(n)}{n^3-n+13}=\frac{\frac{1}{3}+P(n)/n^3}{1-n/n^3+13/n^3}.$

As $P(n)$ is a polynomial of degree 2, the limit $P(n)/n^3 \to 0$ for $n\to \infty$, hence the limit is $1/3$.

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