What is the Limit Comparison Test for Series Convergence?

In summary, the two equations given are a and b, with a represented by 3/n^2 and b represented by 1/n. The limit comparison test is used to compare these equations and it is shown that 3/n^2 is a good comparison for a while 1/n is a good comparison for b. For large values of n, the highest power of n will dominate and can be used to simplify the equations. Using the integral test, it can be shown that a converges while b diverges.
  • #1
rcmango
234
0

Homework Statement



heres the equations: http://img407.imageshack.us/img407/738/untitledzk8.jpg [Broken]

there are two equations, a and b.

Homework Equations



3/n^2 for a,

and 1/n for b.

The Attempt at a Solution



Using the limit comparison test, i don't understand why to use 3/n^2 and 1/n for these equations.

It was explained that we use the largest n in the numerator and denominator, but what we picked, clearly isn't the largest?

please help. thanks.

also, how does a converge, and b diverge??
 
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  • #2
We use 3/n^2 because its a good comparison, a is similar to it. Same for b, if you don't include the 19n, it simplifies to 1/n.

Now, 3/n^2 fulfils all the requirements to converge. The terms converge to zero etc. It can be shown that it is equal to [itex]\frac{\pi^2}{2}[/itex], thought I won't post the proof here.

1/n is the harmonic series, represented by zeta(1) or a p series of p=1.

A p series is of the form [tex]\sum_{n=1}^{\inf} \frac{1}{n^p}[/tex]. The rule of convergence of these is they converge if p>1, or diverge < or =1.
 
  • #3
rcmango said:

Homework Statement



heres the equations: http://img407.imageshack.us/img407/738/untitledzk8.jpg [Broken]

there are two equations, a and b.

Homework Equations



3/n^2 for a,

and 1/n for b.

The Attempt at a Solution



Using the limit comparison test, i don't understand why to use 3/n^2 and 1/n for these equations.

It was explained that we use the largest n in the numerator and denominator, but what we picked, clearly isn't the largest?

please help. thanks.

also, how does a converge, and b diverge??
For large values of n, the highest power of n will "dominate" (100000003 if far larger than 1000000002 [exactly 100000000 times as large!]) so we can ignore lower powers: In a, the numerator is "dominated" by 3n while the denominator is "dominated" by n3. For very large n, the fraction will be close to [tex]\frac{3n}{n^3}= \frac{3}{n^2}[/tex].

In b, n2+ 19n is "dominated by n2 and the square root of that is n. For very large n, [tex]\frac{1}{\sqrt{n^2+ 19n}}[/itex] will be close to [tex]\frac{1}{n}[/tex].

As for why one converges and the other diverges, use the "integral test":
[tex]\int \frac{3}{x^2}dx= -\frac{3}{x}[/tex]
which goes to 0 as x goes to infinity.
On the other hand,
[tex]\int \frac{1}{x}dx= ln(x)[/itex]
which does not converge as x goes to infinity.
 
Last edited by a moderator:
  • #4
awesome job, halls thanks for showing the simplification too. i see it now. thanks.
 

1. What is the purpose of the limit comparison test?

The purpose of the limit comparison test is to determine the convergence or divergence of a given series by comparing it to a known series with known convergence or divergence.

2. How does the limit comparison test work?

The limit comparison test involves taking the ratio of the terms of the given series and the known series. If the limit of this ratio is a finite, non-zero number, then the two series have the same convergence or divergence behavior. If the limit is zero, then the given series converges if the known series also converges, and diverges if the known series diverges. If the limit is infinity, then the given series converges if the known series diverges, and vice versa.

3. What is the known series used in the limit comparison test?

The known series used in the limit comparison test is typically a p-series (such as the harmonic series) or a geometric series. These series have well-known convergence or divergence behavior, making them useful for comparison purposes.

4. Can the limit comparison test be used for all series?

No, the limit comparison test can only be used for series with positive terms. It is also not applicable for series with alternating signs, as it requires taking the ratio of terms.

5. What is the difference between the limit comparison test and the ratio test?

The limit comparison test and the ratio test are similar in that they both involve taking the limit of a ratio of terms. However, the ratio test only compares the given series to a geometric series, while the limit comparison test allows for comparison to a wider range of known series with different convergence or divergence behavior. Additionally, the ratio test can sometimes be inconclusive, while the limit comparison test is always conclusive.

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