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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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    +1. Please look at my answer - is that sufficient?2012-11-29
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    @Flavius you seem to be missing some `\sum`s in the $\LaTeX$ of your update.2012-11-29
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    @anon updated. Better?2012-11-29
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    @Flavius: i) When you write something like $a_n\ge b_n\Rightarrow \lim a_n\ge\lim b_n$, you should always add that this is under the assumption that the limits exist. For instance, $a_n=2$ is greater than $b_n=(-1)^n$, and $a_n$ has a limit and $b_n$ doesn't. ii) Assuming that the limits exist, from $a_n\gt b_n$ you can only conclude $\lim a_n\ge\lim b_n$; for instance, both $a_n=1/n$ and $b_n=0$ converge to $0$ though $a_n\gt b_n$. (Even your own conclusion that both limits are positive infinity contradicts that one is greater than the other.)2012-11-29
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    @Flavius: iii) The $\lim$ in front of the series makes no sense; the upper summation limit $\infty$ already implies by convention that the sum is taken to some upper limit $m$ and then the limit $m\to\infty$ is considered.2012-11-29
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    @joriki so how show I address `i)`?2012-11-29
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    Note that $\lim\limits_{n\to\infty}s_n\ge\lim\limits_{n\to\infty}1/n$ is irrelevant to the argument. You want $s_n>1/n$ for all $n$, which implies $\displaystyle \sum_{k=2}^ns_n>\sum_{k=2}^n1/k$ for all $n$, which implies $\displaystyle\sum_{k=2}^\infty s_n\ge\sum_{k=2}^\infty\frac{1}{k}=\infty$. Of course, to make sense of $\ge$ or $=\infty$, we need to understand what these mean in terms of the definition of limits.2012-11-29
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    @Flavius: You don't need a theorem in that generality here; you quite rightly stated in the question that you want to use the comparison test, so you should only make the statements you need for that, and not more general statements about the comparison of limits of convergent sequences. All you need is to formulate somewhat more precisely what you already wrote in the question: "I must find a lower series which diverges, in order to prove that it diverges".2012-11-29
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    @joriki So I shouldn't bother about the sequences behind the series at all? Perhaps just keep them in mind, but not really write it out?2012-11-29
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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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    The ratio test is inconclusive because $|a_{n+1}/a_n| \to 1$.2012-11-29
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    @FlybyNight: I guess he meant ratio comparison test: http://en.wikipedia.org/wiki/Comparison_test#Ratio_comparison_test2012-11-29