Finding the sum of a Series that is converge or Diverge

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In summary, the series ∑ from n=1 to ∞ of (1+2^n)/(3^n) is convergent and its sum is 5/2. To find the sum, we can break up the series into two parts, 1/3^n and 2^n/3^n, and use the Geometric series formula for both terms with n=1 to get the sum of 5/2.
  • #1
Physicsnoob90
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Homework Statement


Determine whether the series is either Converge or Diverge, if it's convergent, find its sum

∑ from n=1 to ∞ of (1+2^n)/(3^n)

Homework Equations


The Attempt at a Solution



Steps:
1) i replaced the (1+2^n) to just (2^n) and my equation behaves like ∑ (2/3)^n which the Radius (2/3) < 1. So it's Convergent.
2) I found that my leading term was 2/3 by n=1
3) i plug everything into my Geometric series formula (2/3)/(1-(2/3), thus giving me an answer of 2.

Problem: the Books answer was 5/2. Can anyone help me out? Thanks!
 
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  • #2
Physicsnoob90 said:

Homework Statement


Determine whether the series is either Converge or Diverge, if it's convergent, find its sum

∑ from n=1 to ∞ of (1+2^n)/(3^n)

Homework Equations





The Attempt at a Solution



Steps:
1) i replaced the (1+2^n) to just (2^n) and my equation behaves like ∑ (2/3)^n which the Radius (2/3) < 1. So it's Convergent.
2) I found that my leading term was 2/3 by n=1
3) i plug everything into my Geometric series formula (2/3)/(1-(2/3), thus giving me an answer of 2.

Problem: the Books answer was 5/2. Can anyone help me out? Thanks!

You can't throw away the 1 if you expect to get an exact sum. Put it back in.
 
  • #3
Yes, keep the 1 inside. I understand your idea for breaking up the sum, but you forgot about the other term.

Your split idea is good: (1+2^n)/(3^n) = 1/3^n + 2^n/3^n.

Use your formula for both these terms. Hint: 1 = 1^n.
 

1. How do you determine if a series is convergent or divergent?

The convergence or divergence of a series can be determined by evaluating the limit of the series as the number of terms approaches infinity. If the limit exists and is a finite number, the series is convergent. If the limit does not exist or is infinite, the series is divergent.

2. What is the difference between absolute and conditional convergence?

Absolute convergence refers to a series in which the individual terms are all positive, and the series converges regardless of the order in which the terms are added. In contrast, conditional convergence occurs when the series converges only if the terms are added in a specific order.

3. How does the ratio test help in determining the convergence of a series?

The ratio test is a method for determining the convergence or divergence of a series by taking the limit of the ratio of consecutive terms. If the limit is less than 1, the series is absolutely convergent. If the limit is greater than 1, the series is divergent. If the limit is exactly 1, the test is inconclusive and another method must be used.

4. Can a series be both convergent and divergent?

No, a series cannot be both convergent and divergent. It can either converge to a finite value or diverge to infinity or negative infinity. If the limit of the series does not exist, it is considered divergent.

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

To find the sum of a convergent series, you can use the formula for the sum of an infinite geometric series, which is a/(1-r), where a is the first term and r is the common ratio. If the series is not geometric, you can use other methods such as partial sums or integration to find the sum.

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