Questions: Determine whether the series sum from n=1 to infinity of (3^n)/(4^(n-1)) is convergent or divergent. If it is convergent, find its sum.

Solution Steps

To determine whether the series \(\sum_{n=1}^{\infty} \frac{3^{n}}{4^{n-1}}\) is convergent or divergent, we can rewrite the general term and identify the type of series. If it is a geometric series, we can use the formula for the sum of an infinite geometric series.

1. Rewrite the general term: \(\frac{3^n}{4^{n-1}} = 3 \cdot \left(\frac{3}{4}\right)^{n-1}\).
2. Identify the common ratio \(r = \frac{3}{4}\).
3. Check if \(|r| < 1\) to determine convergence.
4. If convergent, use the sum formula for an infinite geometric series: \(S = \frac{a}{1 - r}\), where \(a\) is the first term.

We start with the series

\[ \sum_{n=1}^{\infty} \frac{3^{n}}{4^{n-1}}. \]

We can rewrite the general term as

\[ \frac{3^n}{4^{n-1}} = 3 \cdot \left(\frac{3}{4}\right)^{n-1}. \]

The series can be recognized as a geometric series with the first term

\[ a = 3 \]

and the common ratio

\[ r = \frac{3}{4} = 0.75. \]

To determine if the series converges, we check the absolute value of the common ratio:

\[ |r| = 0.75 < 1. \]

Since the condition for convergence of a geometric series is satisfied, the series is convergent.

The sum \(S\) of an infinite geometric series is given by the formula

\[ S = \frac{a}{1 - r}. \]

Substituting the values we have:

\[ S = \frac{3}{1 - \frac{3}{4}} = \frac{3}{\frac{1}{4}} = 3 \cdot 4 = 12. \]

Final Answer