# Math Help - Comparison test

1. ## Comparison test

Use the comparison test to determine whether each of the following series converges or diverges.

$\sum \frac{2n^2+15n}{n^3+7}$
This is harder than it looks. My supervisor and I got stuck on it!

Here's what I tried:

$0< \sum \frac{2n^2+15n}{n^3+7}< \sum \frac{2n^2+15n+7}{n^3+7}$

$= \sum \frac{(2n+1)(n+7)}{n^3+7}= \sum \frac{2n(n+7)}{n^3+7}+ \frac{n+7}{n^3+7}< \sum \frac{2n^2+14n}{n^3}+\sum \frac{n+7}{n^3}$

$=\sum \frac{2}{n}+\sum \frac{14}{n^2}+\sum \frac{1}{n^2} +\sum \frac{7}{n^3}$

This was the closest I got. I have to find a way to get rid of $\sum \frac{2}{n}$ since that's the only one that diverges.

However, i'm assuming that the sequence is convegent since I couldn't get it to diverge by the comparison test either!

2. Hi,
Originally Posted by Showcase_22
However, i'm assuming that the sequence is convegent since I couldn't get it to diverge by the comparison test either!
This series diverges since for $n\geq1$

$\frac{2n^2+15n}{n^3+7}\geq \frac{2n^2+14n}{n^3+7n^2}=\frac{2}{n}$.

3. Originally Posted by Showcase_22
This is harder than it looks. My supervisor and I got stuck on it!

Here's what I tried:

$0< \sum \frac{2n^2+15n}{n^3+7}< \sum \frac{2n^2+15n+7}{n^3+7}$

$= \sum \frac{(2n+1)(n+7)}{n^3+7}= \sum \frac{2n(n+7)}{n^3+7}+ \frac{n+7}{n^3+7}< \sum \frac{2n^2+14n}{n^3}+\sum \frac{n+7}{n^3}$

$=\sum \frac{2}{n}+\sum \frac{14}{n^2}+\sum \frac{1}{n^2} +\sum \frac{7}{n^3}$

This was the closest I got. I have to find a way to get rid of $\sum \frac{2}{n}$ since that's the only one that diverges.

However, i'm assuming that the sequence is convegent since I couldn't get it to diverge by the comparison test either!
Or try the limit comparison test, comparing the the Harmonic Series.

4. What's the limit form of the comparison test?

I thought it would be the same as what i'm already doing except for the addition of "lim" in front of every sum symbol.

=S

5. Originally Posted by Showcase_22
What's the limit form of the comparison test?

I thought it would be the same as what i'm already doing except for the addition of "lim" in front of every sum symbol.

=S
It roughly states: Let $\sum{a_n}$ and $\sum{b_n}$ be two infinite series. Then if $\lim_{n\to\infty}\frac{a_{n}}{b_n}=c\in\mathbb{R^+ }\implies\sum{a_n}\text{ and }\sum{b_n}\text{ share convergence/divergence}$

6. Originally Posted by Mathstud28
It roughly states: Let $\sum{a_n}$ and $\sum{b_n}$ be two infinite series. Then if $\lim_{n\to\infty}\frac{a_{n}}{b_n}=c\in\mathbb{R^+ }\implies\sum{a_n}\text{ and }\sum{b_n}\text{ share convergence/divergence}$
I haven't seen this before but there's something I was wondering:

Your $\Rightarrow$ only goes one way so I can't use the limits of the numerator and the denominator to imply that the fraction will either converge or diverge.

Am I applying the limit form of the comparison test to the wrong sequence?

7. Originally Posted by Showcase_22
I haven't seen this before but there's something I was wondering:

Your $\Rightarrow$ only goes one way so I can't use the limits of the numerator and the denominator to imply that the fraction will either converge or diverge.

Am I applying the limit form of the comparison test to the wrong sequence?
If your asking that if $\sum{a_n}$ and $\sum{b_n}$ share convergence/divergence then this implies that $\lim_{n\to\infty}\frac{a_n}{b_n}=c\in\mathbb{R^+}$. If so the answer is you can, it is just not pertinent.