Basic Geometric Series Question

Thanks for pointing it out!And thanks for the summary! It's a great way to get people to see what they said in a different light or to get them to see if they really want to say it the way they did.
  • #1
Saterial
54
0

Homework Statement


Determine whether the series is convergent or divergent. Find the sum if possible

Ʃ 1+2^n / 3^n n=1 -> infinity


Homework Equations


a/1-r


The Attempt at a Solution



I split it up so that the equation is now:

Ʃ (1/3^n) + (2/3)^n n=1 -> infinity
Ʃ (1/3^n) + Ʃ (2/3)^n n=1 -> infinity

I know that it is convergent in the second term because 2/3 < 1, how do I setup a/1-r for term 1? :S

a1=1/3 a1 = 2/3
(1/3)/(1-(1/3)) + (2/3)/(1-(2/3))
=5/2

The answer should be 3/2 on the answer sheet it says ?
 
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  • #2
Saterial said:

Homework Statement


Determine whether the series is convergent or divergent. Find the sum if possible

Ʃ 1+2^n / 3^n n=1 -> infinity


Homework Equations


a/1-r


The Attempt at a Solution



I split it up so that the equation is now:

Ʃ (1/3^n) + (2/3)^n n=1 -> infinity
Ʃ (1/3^n) + Ʃ (2/3)^n n=1 -> infinity

I know that it is convergent in the second term because 2/3 < 1, how do I setup a/1-r for term 1? :S

a1=1/3 a1 = 2/3
(1/3)/(1-(1/3)) + (2/3)/(1-(2/3))
=5/2

The answer should be 3/2 on the answer sheet it says ?

USE PARENTHESES!

I don't see anything wrong with your answer, but I had a hard time trying to figure out what your problem was.

This is what you wrote (fixed up in LaTeX):
$$ \sum_{n = 1}^{\infty} \left( 1 + \frac{2^n}{3^n}\right)$$

This is what I'm pretty sure you meant:
$$ \sum_{n = 1}^{\infty} \left(\frac{1 + 2^n}{3^n}\right)$$

Don't write 1+2^n / 3^n if you mean (1+2^n )/ 3^n.
 
  • #3
The first sum, once you've broken them up, is not a geometric sum. Think about what that guy's doing for a little bit.
 
  • #4
gustav1139 said:
The first sum, once you've broken them up, is not a geometric sum. Think about what that guy's doing for a little bit.
Sure it is.
The first series can be written as
$$\sum_{n = 1}^{\infty}\left(\frac{1}{3}\right)^n $$

or 1/3 + (1/3)2 + (1/3)3 + ... +
 
  • #5
derp. right.
 
  • #6
gustav1139 said:
The first sum, once you've broken them up, is not a geometric sum. Think about what that guy's doing for a little bit.

Had I been Mark44, I would have issued you a warning or infraction for that last sentence.
 
  • #7
LCKurtz said:
Had I been Mark44, I would have issued you a warning or infraction for that last sentence.

I apologize. Personification is against the rules?
 
Last edited:
  • #8
gustav1139 said:
The first sum, once you've broken them up, is not a geometric sum. Think about what that guy's doing for a little bit.

LCKurtz said:
Had I been Mark44, I would have issued you a warning or infraction for that last sentence.

gustav1139 said:
I apologize. Personification is against the rules?

I'm not the boss around here but I just wanted to alert you that your comment might easily be construed as violating the section below. It struck me that way anyway.

From the Rules menu at the top of the page:

Langauge and Attitude: Foul or hostile language will not be tolerated on Physics Forums. This includes profanity, obscenity, or obvious indecent language; direct personal attacks or insults; snide remarks or phrases that appear to be an attempt to "put down" another member; and other indirect attacks on a member's character or motives.
 
  • #9
Ah. When I said "that guy," I was referring to the first sum in the OP's question. I guess I see how that could be misconstrued.
 

What is a basic geometric series?

A basic geometric series is a sequence of numbers where each term is found by multiplying the previous term by a constant value. This constant value is known as the common ratio. The general form of a geometric series is a + ar + ar^2 + ar^3 + ..., where a is the first term and r is the common ratio.

What is the formula for finding the sum of a basic geometric series?

The formula for finding the sum of a basic geometric series is S = a(1-r^n)/(1-r), where S is the sum, a is the first term, r is the common ratio, and n is the number of terms in the series.

How do you determine if a basic geometric series converges or diverges?

A basic geometric series will converge if the absolute value of the common ratio (|r|) is less than 1. It will diverge if |r| is greater than or equal to 1.

What is the relationship between a basic geometric series and a geometric sequence?

A basic geometric series is the sum of all terms in a geometric sequence. A geometric sequence is a list of numbers where each term is found by multiplying the previous term by a constant value, known as the common ratio.

How do you find the value of a specific term in a basic geometric series?

The formula for finding the value of a specific term in a basic geometric series is given by Tn = ar^(n-1), where Tn is the value of the nth term, a is the first term, and r is the common ratio.

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