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Investigate the series for convergence and if possible, determine its limit: $\sum\limits_{n=2}^\infty\frac{n^3+1}{n^4-1}$

My thoughts

Let there be the sequence $s_n = \frac{n^3+1}{n^4-1}, n \ge 2$.

I have tried different things with no avail. I suspect I must find a lower series which diverges, in order to prove that it diverges, and use the comparison test.

Could you give me some hints as a comment? Then I'll try to update my question, so you can double-check it afterwards.

Update

$s_n \gt \frac{n^3}{n^4} = \frac1n$

which means that

$\lim\limits_{n\to\infty} s_n > \lim\limits_{n\to\infty}\frac1n$

but $\sum\limits_{n=2}^\infty\frac1n = \infty$

so $\sum\limits_{n=2}^\infty s_n = \infty$

thus the series $\sum\limits_{n=2}^\infty s_n$ also diverges.

The question is: is this formally sufficient?

4 Answers 4

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$\frac{n^3+1}{n^4-1}\gt\frac{n^3}{n^4}=\frac1n\;.$

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    @Flavius: I don't understand why you say that. The whole point of the comparison test is to compare the sequences. If you weren't aware of that, check out the Wikipedia article that you linked to; it shows the comparison of the terms of the sequence.2012-11-29
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Use the limit comparison test with the series $1/1+1/2+1/3+...$

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Try the limit comparison test with $\sum_{n=1}^\infty \frac1n$, i.e. calculatte the limit $\lim_{n\to\infty}\frac{\frac{n^3+1}{n^4-1}}{\frac1n}$

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Do you know the ratio test?

In this case, try to compare with $\sum_{n=0}^\infty \frac{1}{n}$.

EDIT. Ok, the reference was wrong. What I meant is the criterion which says: let $\sum a_n$ and $\sum b_n$ be two (positive) series, and assume

$ \mathrm{lim}\ \frac{a_n}{b_n} = L, \qquad \text{with $L \neq 0,\infty$} $

then

$ \sum a_n \quad \text{converges} \qquad \Longleftrightarrow \qquad \sum b_n \quad \text{converges} \ . $

Which in our case, says:

$ \mathrm{lim}\ \frac{\frac{n^3+1}{n^4 -1}}{\frac{1}{n}} = \mathrm{lim}\ \frac{n^4 + n}{n^4 - 1} = 1 \ . $

Hence, as $\sum \frac{1}{n}$ diverges, so does $\sum \frac{n^3+1}{n^4 -1}$.

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    @FlybyNight: I guess he meant ratio comparison test: http://en.wikipedia.org/wiki/Comparison_test#Ratio_comparison_test2012-11-29