Is the Series Ʃ (1 + 2^n) / 3^n Convergent and What is Its Sum?

Click For Summary

Homework Help Overview

The discussion revolves around the convergence of the series Ʃ (1 + 2^n) / 3^n from n=1 to infinity. Participants are exploring whether the series converges or diverges and are attempting to find its sum, if possible.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss splitting the series into two parts: Ʃ (1/3^n) and Ʃ (2/3)^n. There is uncertainty about how to apply the formula a/(1-r) for the first term. Some participants express confusion regarding the nature of the first sum and whether it qualifies as a geometric series.

Discussion Status

There is ongoing exploration of the series' components, with some participants questioning the setup and definitions used. Guidance has been offered regarding the notation and the interpretation of the series, but no consensus has been reached on the convergence or the correct sum.

Contextual Notes

Participants are navigating potential misunderstandings related to the notation of the series and the rules of the forum regarding language and conduct. There is a mention of the answer sheet indicating a different sum than what some participants have calculated.

Saterial
Messages
54
Reaction score
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 ?
 
Physics news on Phys.org
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.
 
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.
 
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 + ... +
 
derp. right.
 
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.
 
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:
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:

language 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.
 
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.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
14
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
Replies
7
Views
2K