Comparison Test: How & Why for Next Step?

In summary, the conversation discusses how to choose the second term for a series comparison problem. The teacher suggests using the limit comparison test, but the student is unsure of how to choose the second term. Another student suggests using a convergent series as a comparison, specifically \sum_{n = 1}^\infty \frac{1}{(3/2)^n}. The conversation ends with the student being asked to finish the problem using this convergent series.
  • #1
wajed
57
0
http://img205.imageshack.us/img205/5117/summation.jpg [Broken]

I think next step is: http://img205.imageshack.us/img205/7044/summation2.jpg [Broken]

but the questions are: how & why?

why is that the next step?
how do I solve such problems?\how do I choose the second part?
 
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  • #2
For sure we are comparing so there is a "less than" sign..
I`m just concerned on how to choose the terms on the left?
 
  • #3
For the series
[tex]\sum_{n = 1}^\infty \frac{1}{2^n - 1}[/tex]

the obvious choice would be a comparison with the convergent series
[tex]\sum_{n = 1}^\infty \frac{1}{2^n}[/tex]
Unfortunately, all of the terms in the first series are larger than those of the convergent series, so that comparison won't do any good.

I have no idea what you're trying to do with this series:
[tex]\sum_{n = 1}^\infty \frac{1}{2^n - 1^{n - 1}}[/tex]

Do you know the limit comparison test?
 
  • #4
The teacher have just talked about it today,
and he talked about the comparison test first..

This problem, he solved it before talking about LCT..
So, its solved by CT..
Now why is it solved like that, or specifically, why is that second part chosen to be like that, the teacher himself said that he doesn't know!
Well, he wasn't going to say it.. but someone asked (and I was going to ask) about that step.. then he said "well, its just that you choose it like that, this is how the problem is solved.."

So I`m not here seeking for a final answer.. the question is: how do I choose the 2nd parts.. and why aren`t you, nor my teacher, able to know how to choose the 2nd term?
(what are even the steps that I should take to put/know the second term?)
 
  • #5
OK, here's a convergent series you can use for comparison:
[tex]\sum_{n = 1}^\infty \frac{1}{(3/2)^n}[/tex]

2n - 1 > (3/2)n for n > 1
so 1/(2n - 1) < 1/(3/2)n for n > 1

Can you finish it?
 

1. How does the Comparison Test work?

The Comparison Test is a method used to determine the convergence or divergence of an infinite series. It involves comparing the given series to a known series whose convergence or divergence is already known. If the known series converges, then the given series also converges. If the known series diverges, then the given series also diverges.

2. What is the purpose of using the Comparison Test?

The Comparison Test is used to determine the convergence or divergence of an infinite series that does not have a known closed form solution. It allows us to use the convergence or divergence of a known series to determine the convergence or divergence of a more complicated series.

3. Does the Comparison Test always work?

No, the Comparison Test does not always work. It is only applicable to series whose terms are positive and decreasing. If the terms of the given series are not positive and decreasing, then the Comparison Test cannot be used.

4. How do you choose the appropriate series to compare to?

The series chosen for comparison should have similar properties to the given series. This means that they should have the same behavior as the given series as n approaches infinity. It is also important to choose a known series whose convergence or divergence is already known.

5. Can the Comparison Test be used to determine the exact sum of a series?

No, the Comparison Test only tells us whether a series converges or diverges. It does not provide an exact value for the sum of the series. To find the exact sum, other methods such as the Ratio Test or the Integral Test may be used.

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