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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?

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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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  • $\begingroup$ +1. Please look at my answer - is that sufficient? $\endgroup$
    – Math for fun
    Commented Nov 29, 2012 at 19:31
  • $\begingroup$ @Flavius you seem to be missing some \sums in the $\LaTeX$ of your update. $\endgroup$
    – anon
    Commented Nov 29, 2012 at 19:34
  • $\begingroup$ @anon updated. Better? $\endgroup$
    – Math for fun
    Commented Nov 29, 2012 at 19:36
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    $\begingroup$ @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.) $\endgroup$
    – joriki
    Commented Nov 29, 2012 at 19:37
  • $\begingroup$ @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. $\endgroup$
    – joriki
    Commented Nov 29, 2012 at 19:38
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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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