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Geometric series question

  • Thread starter MarcL
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  • #1
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Homework Statement



Determine whether the serie is convergent or divergent , if it is convergent find its sum.

Ʃn=1 (1 + 2n )/ 3n

Homework Equations



Ʃa(r)n-1 = a / (1-r) r < 1 is converging or if r > 1 diverging

The Attempt at a Solution



Well I can see its a geometric series so i did Ʃ (1/3)^n + (2/3)^n and r < 1 therefore its convergent but I can't figure out the a. I thought it would be one but the answer key says 1/3 and 2/3 as the a for both series
 

Answers and Replies

  • #2
Dick
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Homework Statement



Determine whether the serie is convergent or divergent , if it is convergent find its sum.

Ʃn=1 (1 + 2n )/ 3n

Homework Equations



Ʃa(r)n-1 = a / (1-r) r < 1 is converging or if r > 1 diverging

The Attempt at a Solution



Well I can see its a geometric series so i did Ʃ (1/3)^n + (2/3)^n and r < 1 therefore its convergent but I can't figure out the a. I thought it would be one but the answer key says 1/3 and 2/3 as the a for both series
Find the sum for (1/3)^n and (2/3)^n separately and then add them. They don't have to have a common value of a. The whole series isn't geometric. Only the parts are.
 
  • #3
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Oh I know I mean, the a value for (1/3)^n is said to be 1/3 while the one for (2/3)^n is said to be 2/3. I don't understand how the this came to be. I thought a in each case was 1
 
  • #4
Dick
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Oh I know I mean, the a value for (1/3)^n is said to be 1/3 while the one for (2/3)^n is said to be 2/3. I don't understand how the this came to be. I thought a in each case was 1
Your formula in the "Relevant equations" section isn't right or wrong until you put some limits on the summation like you did in the problem statement. What are they?
 
  • #5
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oh sorry its n=1 to infinity. My bad I hadn't notice that!
 
  • #6
Dick
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oh sorry its n=1 to infinity. My bad I hadn't notice that!
Then that makes the a's what the problem solution suggests, right?
 
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  • #7
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I feel so dumb.... I was so use to n-1 I assumed right away r = 1 due to the exponent for the first term. Dear, let's hope rest will make me smarter for my final. Thanks a lot Dick for bearing through this question haha!
 
  • #8
Dick
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I feel so dumb.... I was so use to n-1 I assumed right away r = 1 due to the exponent for the first term. Dear, let's hope rest will make me smarter for my final. Thanks a lot Dick for bearing through this question haha!
You are pretty smart if you figured that out so quickly given the hint. You'll do fine. Just pay attention to details. Like limits on summations.
 
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  • #9
Ray Vickson
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Homework Statement



Determine whether the serie is convergent or divergent , if it is convergent find its sum.

Ʃn=1 (1 + 2n )/ 3n

Homework Equations



Ʃa(r)n-1 = a / (1-r) r < 1 is converging or if r > 1 diverging

The Attempt at a Solution



Well I can see its a geometric series so i did Ʃ (1/3)^n + (2/3)^n and r < 1 therefore its convergent but I can't figure out the a. I thought it would be one but the answer key says 1/3 and 2/3 as the a for both series
Don't forget that an infinite series is defined as the limit of a finite sum, so
[tex] S = \sum_{n=1}^{\infty} \frac{1+2^n}{3^n} [/tex]
is defined as
[tex] S = \lim_{N \to \infty} \sum_{n=1}^{N} \frac{1+2^n}{3^n} [/tex]
For finite N the sum can be split into two obvious parts, and both parts converge separately as ##N \to \infty.## It follows that the original infinite series is the sum of the two separate series, each of which is a convergent geometric series.
 

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