How to Show the Comparison Inequality for a Series

[Mutaja]

Homework Statement

Use the comparison test to show that the series converges, and find the value it converge to by using partial fractions.

∑ n=1 -> ∞: [itex]\frac{2}{n^2 + 5n + 6}[/itex]

Homework Equations

The Attempt at a Solution

The series can be written as 2 * ∑ n=1 -> ∞: [itex]\frac{1}{n^2 + 5n + 6}[/itex]

Since 5n + 6 is neglectible:

2 * ∑ n=1 -> ∞: [itex]\frac{1}{n^2}[/itex]. This series will converge.

Therefore we can guess that 2* ∑ n=1 -> ∞: [itex]\frac{1}{n^2 + 5n + 6}[/itex] will converge.

Now I have to find a larger series which also converges.

[itex]\frac{1}{n^2 + 5n + 6}[/itex] < [itex]\frac{1}{n^2 + 5n}[/itex]

Now I'm supposed to rewrite [itex]\frac{1}{n^2 + 5n}[/itex] as something + something. This is where I'm stuck.

Are anyone familiar with this method, and can point me into the right direction?

Any feedback will as always be appreciated.

Thanks.

 

[STEMucator]

You did your comparison test properly. ##\frac{1}{n^2}## converges so you know your original series will converge.

To use partial fractions, write: ##n^2 + 5n + 6 = (n+2)(n+3)##.

 

[Mutaja]

You did your comparison test properly. ##\frac{1}{n^2}## converges so you know your original series will converge.

To use partial fractions, write: ##n^2 + 5n + 6 = (n+2)(n+3)##.

Alright, thank you.

So, I've rewritten the series as:

[itex]\frac{1}{n+2}[/itex] - [itex]\frac{1}{n+3}[/itex].

Adding the series then gives [itex]\frac{1}{3}[/itex] - [itex]\frac{1}{n+3}[/itex]

Since the limit of [itex]\frac{1}{3}[/itex] - [itex]\frac{1}{n+3}[/itex] = [itex]\frac{1}{3}[/itex]

The sum which the series converge to is [itex]\frac{1}{3}[/itex].

It looks correct according to my notes at least.

Thanks so much for helping me out.

 

[STEMucator]

Alright, thank you.

So, I've rewritten the series as:

[itex]\frac{1}{n+2}[/itex] - [itex]\frac{1}{n+3}[/itex].

Adding the series then gives [itex]\frac{1}{3}[/itex] - [itex]\frac{1}{n+3}[/itex]

Since the limit of [itex]\frac{1}{3}[/itex] - [itex]\frac{1}{n+3}[/itex] = [itex]\frac{1}{3}[/itex]

The sum which the series converge to is [itex]\frac{1}{3}[/itex].

It looks correct according to my notes at least.

Thanks so much for helping me out.

No not quite. We have:

##n^2 + 5n + 6 = (n+2)(n+3)##.

So that:

##\frac{2}{n^2 + 5n + 6} = \frac{A}{n+2} + \frac{B}{n+3}##

Where the relationship:

##2 = A(n+3) + B(n+2)## must hold.

If ##n = -3##, then ##B = -2##. If ##n = -2##, then ##A = 2##.

So really you have ##\frac{2}{n+2} - \frac{2}{n+3}##

You were missing a factor of ##2##.

 

[Mutaja]

You were missing a factor of ##2##.

Indeed, my bad. I didn't consider the fact that I had taken it outside of the actual series, and I forgot to multiply it back into get my final answer. I should be 2 * 1/3 = 2/3.

Now, it should be good :)

 

[Mark44]

Homework Statement

Use the comparison test to show that the series converges, and find the value it converge to by using partial fractions.

∑ n=1 -> ∞: [itex]\frac{2}{n^2 + 5n + 6}[/itex]

Just a little bit more LaTeX will make the summation look a lot nicer.

$$\sum_{n = 1}^\infty \frac{2}{n^2 + 5n + 6}$$

The above renders as
$$\sum_{n = 1}^\infty \frac{2}{n^2 + 5n + 6}$$

Homework Equations

The Attempt at a Solution

The series can be written as 2 * ∑ n=1 -> ∞: [itex]\frac{1}{n^2 + 5n + 6}[/itex]

Since 5n + 6 is neglectible:

2 * ∑ n=1 -> ∞: [itex]\frac{1}{n^2}[/itex]. This series will converge.

Therefore we can guess that 2* ∑ n=1 -> ∞: [itex]\frac{1}{n^2 + 5n + 6}[/itex] will converge.

Now I have to find a larger series which also converges.

[itex]\frac{1}{n^2 + 5n + 6}[/itex] < [itex]\frac{1}{n^2 + 5n}[/itex]

It's more convincing to compare directly with ##\sum \frac{1}{n^2}##, a convergent p-series.
##\sum \frac{1}{n^2}## is a convergent series, so ##2\sum \frac{1}{n^2}## is, as well.
It's not hard to show that ##\frac 1 {n^2 + 5n + 6} < \frac 1 {n^2}##.

Now I'm supposed to rewrite [itex]\frac{1}{n^2 + 5n}[/itex] as something + something. This is where I'm stuck.

Are anyone familiar with this method, and can point me into the right direction?

Any feedback will as always be appreciated.

Thanks.

 

[Mutaja]

Just a little bit more LaTeX will make the summation look a lot nicer.

$$\sum_{n = 1}^\infty \frac{2}{n^2 + 5n + 6}$$

The above renders as
$$\sum_{n = 1}^\infty \frac{2}{n^2 + 5n + 6}$$
It's more convincing to compare directly with ##\sum \frac{1}{n^2}##, a convergent p-series.
##\sum \frac{1}{n^2}## is a convergent series, so ##2\sum \frac{1}{n^2}## is, as well.
It's not hard to show that ##\frac 1 {n^2 + 5n + 6} < \frac 1 {n^2}##.

Thanks for the latex code.

Under normal circumstances, in Calculus 2 (not the actual course, but a equivalent course), would it be needed to actually show that ##\frac 1 {n^2 + 5n + 6} < \frac 1 {n^2}##? It like 2 < 4. I can't show it or prove it.

or do you mean that ##\frac 1 {n^2 + 5n + 6} < \frac 1 {n^2}## is adequate?

[itex]\frac{1}{n^2}[/itex] is convergent, 2 * [itex]\frac{1}{n^2}[/itex] is convergent, therefore ##\frac 1 {n^2 + 5n + 6}## is convergent because ##\frac 1 {n^2 + 5n + 6} < \frac 1 {n^2}##.

 

[STEMucator]

Thanks for the latex code.

Under normal circumstances, in Calculus 2 (not the actual course, but a equivalent course), would it be needed to actually show that ##\frac 1 {n^2 + 5n + 6} < \frac 1 {n^2}##? It like 2 < 4. I can't show it or prove it.

or do you mean that ##\frac 1 {n^2 + 5n + 6} < \frac 1 {n^2}## is adequate?

[itex]\frac{1}{n^2}[/itex] is convergent, 2 * [itex]\frac{1}{n^2}[/itex] is convergent, therefore ##\frac 1 {n^2 + 5n + 6}## is convergent because ##\frac 1 {n^2 + 5n + 6} < \frac 1 {n^2}##.

It is adequate to write ##\frac 1 {n^2 + 5n + 6} < \frac 1 {n^2}##. Although it is easy to show this relationship holds.

For ##n ≥ 1## it is true that ##n^2 + 5n + 6 > n^2##. Reciprocating the inequality, we obtain ##\frac{1}{n^2 + 5n + 6} < \frac{1}{n^2}##.

 

[Mark44]

Whether it's adequate to just write ##\frac 1 {n^2 + 5n + 6} < \frac 1 {n^2}## probably depends on your instructor, and Zondrina's reply shows that it's easy to demonstrate this inequality, working with the reciprocals.

If your instructor is confident in your ability to justify the inequality, he/she might not require that the work be shown. However, it doesn't hurt to get some practice doing these quick proofs, especially if the problem happened to be slightly more difficult, like this one:
$$\sum_{n = 1}^\infty \frac 1 {n^2 - n - 6}$$

The series to compare to is again ##\sum \frac 1 {n^2}##, but this time it's a little harder to show that ##frac{1}{n^2 - n - 6} < \frac 1 {n^2}##.