Finding sum to convergent series?

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Homework Help Overview

The discussion revolves around determining the convergence of the series Ʃ (3^n + 2^n)/6^n and finding its sum if it converges. The subject area is series convergence, particularly geometric series.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to identify the common ratio "r" for the series without substituting values for n. They express confusion about the process of finding "r" and calculating the sum. Some participants suggest breaking the series into two separate sums to analyze convergence individually. Others propose rewriting parts of the series to identify it as a geometric series.

Discussion Status

The discussion is ongoing, with participants exploring different methods to approach the problem. Some guidance has been offered regarding the breakdown of the series and the identification of the geometric series form, but there is no explicit consensus on the final sum or the interpretation of the contributions from each part of the series.

Contextual Notes

Participants are navigating the definitions and properties of geometric series, with some uncertainty about the significance of the terms involved and how they affect convergence. There is a focus on understanding the implications of the series' components rather than resolving the problem outright.

MySecretAlias
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Homework Statement


Decide whether convergent or divergent, if convergent, find sum.


Ʃ as n = 1 and goes to infinity > (3^n + 2^n)/6^n

Homework Equations



a/1-r

The Attempt at a Solution



I'm just confused where to find the "r" to this without actually plugging in values for n, then dividing. For example, to get the answer (at least I think it's correct), I plugged 1 in for n, got 5/6, then i plugged in 2 for n, and got 11/36. I did (11/36) / (5/6), and used that for my r value, plugged that into a/1-r, and got about 1.31

How exactly do I find the r value? Thanks.
 
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Try breaking it into two sums,
[tex]\sum \frac{3^n}{6^n}+ \sum \frac{2^n}{6^n}[/tex]
If those converge, separately, then the combination converges to the sum of those two.
 
I respect that. Thank you. However, how do i find what the sum is?
 
MySecretAlias said:
I respect that. Thank you. However, how do i find what the sum is?

Write 3^n/6^n as (3/6)^n=(1/2)^n. It's a geometric series. You've probably covered those.
 
Does this mean that r value is simply 1/2? the 2^n does not really matter, since in the long run, it becomes so miniscule?
 
MySecretAlias said:
Does this mean that r value is simply 1/2? the 2^n does not really matter, since in the long run, it becomes so miniscule?

Of course the 2 matters, otherwise it wouldn't be (1/2)^n. What's the sum for n=1 to infinity of (1/2)^n? Review geometric series if you have to.
 

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