Finding sum of convergent series.

In summary: But for a quick summary, you can use the alternating series test to show that the series converges. The Leibnitz criterion can be used to show that the series converges absolutely. Also, the identity (cos a)^3 = (1/4)(3cos a + cos 3a) can be used to simplify the series and make it easier to determine convergence.
  • #1
peripatein
880
0
Hi,

I have determined, correctly I believe, that the following series converges:
1/[(3n-2)(3n+1)]
Now I am asked to determine its sum. I have tried separating it into two subseries, but each time got a p-series with p=1, hence to no avail.
The answer should be 1/3, but how may it be arrived at?
 
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  • #2
peripatein said:
Hi,

I have determined, correctly I believe, that the following series converges:
1/[(3n-2)(3n+1)]
Now I am asked to determine its sum. I have tried separating it into two subseries, but each time got a p-series with p=1, hence to no avail.
The answer should be 1/3, but how may it be arrived at?

I would rewrite 1/[(3n-2)(3n+1)] as the sum of two fractions, using partial fraction decomposition. When you have done that, expand the new series. You'll probably find that the series is a telescoping one.
 
  • #3
peripatein said:
Hi,

I have determined, correctly I believe, that the following series converges:
1/[(3n-2)(3n+1)]
Now I am asked to determine its sum. I have tried separating it into two subseries, but each time got a p-series with p=1, hence to no avail.
The answer should be 1/3, but how may it be arrived at?

Partial fractions, following which you get a very simple telescoping series. Write out the first few terms of each.
 
  • #4
EDIT : Whoops i didn't read the whole post, my bad.
 
Last edited:
  • #5
Zondrina said:
An easier way would be to use the comparison test. You should see the denominator is easily dominated by n2
The OP has already determined that the series converges. Now he/she needs to determine its sum.
 
  • #6
Thanks!
How may I know whether the series ((-1)^n)[cos (3^n)x]^3/(3^n) converges/diverges?Should I use the Leibnitz Criterion?
It is stated that (cos a)^3 = (1/4)(3cos a + cos 3a)
 
  • #7
peripatein said:
Thanks!
How may I know whether the series ((-1)^n)[cos (3^n)x]^3/(3^n) converges/diverges?Should I use the Leibnitz Criterion?
It is stated that (cos a)^3 = (1/4)(3cos a + cos 3a)
I suggest starting a new thread for this.
 

1. What is a convergent series?

A convergent series is a mathematical series in which the terms of the series approach a finite limit as the number of terms increases. This means that the sum of the series can be calculated and will not approach infinity as the number of terms increases.

2. How do you find the sum of a convergent series?

To find the sum of a convergent series, you can use a mathematical formula called the sum of a geometric series. This formula is S = a / (1-r), where S is the sum, a is the first term in the series, and r is the common ratio between consecutive terms in the series.

3. Can all series be summed using this method?

No, this method can only be used for convergent series. Divergent series, which have no finite limit, cannot be summed using this formula.

4. What are some examples of convergent series?

Some examples of convergent series include the sum of the infinite geometric series 1/2 + 1/4 + 1/8 + ... which has a sum of 1, and the sum of the infinite harmonic series 1 + 1/2 + 1/3 + ... which has a sum of infinity.

5. Why is finding the sum of convergent series important?

Finding the sum of convergent series is important in mathematics and science because it allows us to calculate the total value of a series without having to add an infinite number of terms. This can be useful in many real-world applications, such as calculating interest or growth rates.

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