Is the Series Convergent or Divergent?

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Homework Help Overview

The discussion revolves around determining the convergence or divergence of a series, specifically the series represented by the expression \(\sum \frac{3+7n}{6n}\). Participants are exploring various methods to analyze the series, including the comparison test and limit comparison test.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the comparison test and its application, with one participant attempting to compare the series to a geometric series. Others question the validity of inequalities used in the comparisons and explore the implications of terms converging to zero.

Discussion Status

The discussion is ongoing, with participants providing hints and questioning each other's reasoning. Some guidance has been offered regarding the conditions for convergence, but there is no explicit consensus on the series' behavior yet.

Contextual Notes

There are indications of confusion regarding the behavior of the terms in the series as \(n\) approaches infinity, particularly concerning the nature of geometric series and their convergence criteria.

tnutty
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Homework Statement



Determine whether the series converges or diverges.

\sum 3+7n / 6n

Attempt :

Comparison test :

3+7n / 6n < 7n / 6n

3+7n / 6n < (6/7)n

since (6/7)n is a geometric series and is convergent is
3+7n / 6n convergent as well?
 
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I see my mistake

3+7n / 6n > (7/6)^n

but then what?
 
Hint: For a series
\sum_{n=0}^{\infty}a_n

to converge, the terms have to converge to zero, i.e.

\lim_{n-&gt;\infty}a_n=0.
 
I could use the limit comparison test but I get stuck.
 
You have a_n=3+7^n/6^n&gt;3, so can a_n converge to zero?
 
How did you figure that inequality ?

(7/6)^n converges to -7
 
tnutty said:
How did you figure that inequality ?

Clearly, (7/6)^n is a positive number for any n.

(7/6)^n converges to -7

I don't think so, 7/6>1, so (7/6)^n goes to infinity for large n.
 
(7/6)^n

This is a geometric series. And I know that by definition a*r^(n-1) ,where |r|<1 = a/(1-r).

so (7/6)^n
=

(7/6) * (7/6)^(n-1)

so,
a = 7/6
r = 7/6

it follows that

(7/6)^n = a/(1-r) = (7/6) / (1-7/6) = -7
 
wait I see what your saying. How comes this the above statement is wrong?
 
  • #10
no your right, 7/6 > 1 so this series diverges.
Thanks
 

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