Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

4 Answers

up vote 6 down vote accepted

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

share|improve this answer
    
+1. Please look at my answer - is that sufficient? –  Flavius Nov 29 '12 at 19:31
    
@Flavius you seem to be missing some \sums in the $\LaTeX$ of your update. –  anon Nov 29 '12 at 19:34
    
@anon updated. Better? –  Flavius Nov 29 '12 at 19:36
    
@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.) –  joriki Nov 29 '12 at 19:37
    
@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. –  joriki Nov 29 '12 at 19:38
show 5 more comments

Use the limit comparison test with the series $1/1+1/2+1/3+...$

share|improve this answer
add comment

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

share|improve this answer
add comment

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 $\sum \frac{n^3+1}{n^4 -1}$.

share|improve this answer
1  
The ratio test is inconclusive because $|a_{n+1}/a_n| \to 1$. –  Fly by Night Nov 29 '12 at 19:20
    
@FlybyNight: I guess he meant ratio comparison test: en.wikipedia.org/wiki/Comparison_test#Ratio_comparison_test –  Lior B-S Nov 29 '12 at 20:35
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.