Answer :
To determine whether the series [tex]\(\sum_{n=1}^{\infty} \frac{6^n}{3^n - 2}\)[/tex] converges or diverges, let's utilize the Comparison Test by comparing it to a simpler geometric series. Here's a detailed step-by-step solution:
1. Rewrite the general term:
Consider the general term of the series:
[tex]\[ a_n = \frac{6^n}{3^n - 2} \][/tex]
2. Find a simpler expression for comparison:
To simplify our comparison, note that for large [tex]\(n\)[/tex], the term [tex]\(3^n\)[/tex] in the denominator will dominate the constant [tex]\(2\)[/tex]. Therefore, [tex]\(\frac{6^n}{3^n - 2}\)[/tex] behaves similarly to [tex]\(\frac{6^n}{3^n}\)[/tex] when [tex]\(n\)[/tex] is large.
Simplifying [tex]\(\frac{6^n}{3^n}\)[/tex]:
[tex]\[ \frac{6^n}{3^n} = \left( \frac{6}{3} \right)^n = 2^n \][/tex]
3. Consider a geometric series for comparison:
The term [tex]\(2^n\)[/tex] suggests that we can compare our original series with the geometric series [tex]\(\sum_{n=1}^{\infty} 2^n \)[/tex].
4. Analyze the comparison:
For large [tex]\(n\)[/tex], we have:
[tex]\[ \frac{6^n}{3^n - 2} \approx \frac{6^n}{3^n} = 2^n \][/tex]
So, the terms of the given series are approximately equal to the terms of the geometric series [tex]\(\sum_{n=1}^{\infty} 2^n\)[/tex].
5. Examine the geometric series [tex]\(\sum_{n=1}^{\infty} 2^n\)[/tex]:
The geometric series [tex]\(\sum_{n=1}^{\infty} ar^n\)[/tex] converges if [tex]\(|r| < 1\)[/tex]. In this case, the series [tex]\(\sum_{n=1}^{\infty} 2^n\)[/tex] has [tex]\(r = 2\)[/tex], which is greater than 1.
Since [tex]\(|2| > 1\)[/tex], the geometric series [tex]\(\sum_{n=1}^{\infty} 2^n\)[/tex] diverges.
6. Comparison Test Conclusion:
Since the terms of our original series [tex]\(\frac{6^n}{3^n - 2}\)[/tex] are eventually larger than or comparable to the terms of the divergent geometric series [tex]\(\sum_{n=1}^{\infty} 2^n\)[/tex], the original series must also diverge.
Therefore, the correct answer is:
The series diverges.
1. Rewrite the general term:
Consider the general term of the series:
[tex]\[ a_n = \frac{6^n}{3^n - 2} \][/tex]
2. Find a simpler expression for comparison:
To simplify our comparison, note that for large [tex]\(n\)[/tex], the term [tex]\(3^n\)[/tex] in the denominator will dominate the constant [tex]\(2\)[/tex]. Therefore, [tex]\(\frac{6^n}{3^n - 2}\)[/tex] behaves similarly to [tex]\(\frac{6^n}{3^n}\)[/tex] when [tex]\(n\)[/tex] is large.
Simplifying [tex]\(\frac{6^n}{3^n}\)[/tex]:
[tex]\[ \frac{6^n}{3^n} = \left( \frac{6}{3} \right)^n = 2^n \][/tex]
3. Consider a geometric series for comparison:
The term [tex]\(2^n\)[/tex] suggests that we can compare our original series with the geometric series [tex]\(\sum_{n=1}^{\infty} 2^n \)[/tex].
4. Analyze the comparison:
For large [tex]\(n\)[/tex], we have:
[tex]\[ \frac{6^n}{3^n - 2} \approx \frac{6^n}{3^n} = 2^n \][/tex]
So, the terms of the given series are approximately equal to the terms of the geometric series [tex]\(\sum_{n=1}^{\infty} 2^n\)[/tex].
5. Examine the geometric series [tex]\(\sum_{n=1}^{\infty} 2^n\)[/tex]:
The geometric series [tex]\(\sum_{n=1}^{\infty} ar^n\)[/tex] converges if [tex]\(|r| < 1\)[/tex]. In this case, the series [tex]\(\sum_{n=1}^{\infty} 2^n\)[/tex] has [tex]\(r = 2\)[/tex], which is greater than 1.
Since [tex]\(|2| > 1\)[/tex], the geometric series [tex]\(\sum_{n=1}^{\infty} 2^n\)[/tex] diverges.
6. Comparison Test Conclusion:
Since the terms of our original series [tex]\(\frac{6^n}{3^n - 2}\)[/tex] are eventually larger than or comparable to the terms of the divergent geometric series [tex]\(\sum_{n=1}^{\infty} 2^n\)[/tex], the original series must also diverge.
Therefore, the correct answer is:
The series diverges.
