To determine whether the series

$\sum_{n=1}^{\infty} \frac{3x - 2}{n 3^n}$

is absolutely convergent or divergent using the Ratio Test, let's proceed with the following steps.

Ratio Test Review

For a series $\sum a_n$, the Ratio Test states:

• If $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L$,
  • If $L < 1$, the series converges absolutely.
  • If $L > 1$, the series diverges.
  • If $L = 1$, the test is inconclusive.

In this case, the $n$-th term $a_n$ of the series is: $a_n = \frac{3x - 2}{n 3^n}.$

Applying the Ratio Test

1. Compute $a_{n+1}$: $a_{n+1} = \frac{3x - 2}{(n+1) 3^{n+1}}.$

2. Calculate $\left| \frac{a_{n+1}}{a_n} \right|$: $\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{\frac{3x - 2}{(n+1) 3^{n+1}}}{\frac{3x - 2}{n 3^n}} \right| = \left| \frac{3x - 2}{3x - 2} \right| \cdot \frac{n}{n+1} \cdot \frac{1}{3}.$ Simplifying, we get: $\left| \frac{a_{n+1}}{a_n} \right| = \frac{n}{n+1} \cdot \frac{1}{3}.$

3. Take the limit as $n \to \infty$: $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{n}{n+1} \cdot \frac{1}{3} = \frac{1}{3}.$

Conclusion

Since $\frac{1}{3} < 1$, the series $\sum_{n=1}^{\infty} \frac{3x - 2}{n 3^n}$ converges absolutely by the Ratio Test.

Would you like further details or have any questions on this solution?

Related Questions

1. How does the Ratio Test compare to the Root Test for determining convergence?
2. Can the Ratio Test be used to determine the radius of convergence of a power series?
3. What would the outcome be if the limit was equal to 1 in the Ratio Test?
4. How would the series behave if the term $3^n$ in the denominator were replaced with a different base, like $4^n$?
5. How does absolute convergence differ from conditional convergence in series?

Tip

When using the Ratio Test, simplifying the terms before taking limits can make calculations more straightforward, especially for terms involving powers and factorials.