Given the series $$ \sum_{n=0}^{\infty} \frac{3^{n}}{3^{3 n}} $$ answer the following questions. Answer Attempt 2 out of 2 $$ \sum_{n=0}^{\infty} \frac{3^{n}}{3^{3 n}} $$ is $$ r=\square \frac{1}{9} \cdot \sum_{n=0}^{\infty} \frac{3^{n}}{3^{3 n}} $$ converges $$ \sim $$ since a geometric series will converge if and only if $$ |r|

Question

Given the series $$ \sum_{n=0}^{\infty} \frac{3^{n}}{3^{3 n}} $$ answer the following questions. Answer Attempt 2 out of 2 $$ \sum_{n=0}^{\infty} \frac{3^{n}}{3^{3 n}} $$ is $$ r=\square \frac{1}{9} \cdot \sum_{n=0}^{\infty} \frac{3^{n}}{3^{3 n}} $$ converges $$ \sim $$ since a geometric series will converge if and only if $$ |r|<1 \quad $$. The sum of the series is with the common ratio .

UpStudy AI · Accepted Answer

We start with the series

$$
\sum_{n=0}^{\infty} \frac{3^n}{3^{3n}}.
$$

Notice that the term inside the series can be simplified as follows:

$$
\frac{3^n}{3^{3n}} = \frac{3^n}{(3^3)^n} = \frac{3^n}{27^n} = \left(\frac{3}{27}\right)^n = \left(\frac{1}{9}\right)^n.
$$

Thus, the series becomes

$$
\sum_{n=0}^{\infty} \left(\frac{1}{9}\right)^n.
$$

This is a geometric series with common ratio

$$
r = \frac{1}{9}.
$$

A geometric series converges if and only if $$|r| < 1$$. Since

$$
\left|\frac{1}{9}\right| = \frac{1}{9} < 1,
$$

the series converges.

The sum $$ S $$ of an infinite geometric series is given by

$$
S = \frac{a}{1 - r},
$$

where $$ a $$ is the first term. In this series, the first term is when $$ n = 0 $$, so

$$
a = \left(\frac{1}{9}\right)^0 = 1.
$$

Thus, the sum of the series is

$$
S = \frac{1}{1-\frac{1}{9}} = \frac{1}{\frac{8}{9}} = \frac{9}{8}.
$$

In summary:

- The common ratio is $$ r = \frac{1}{9} $$.
- The series converges since $$ \left|\frac{1}{9}\right| < 1 $$.
- The sum of the series is $$ \frac{9}{8} $$.
The series $$ \sum_{n=0}^{\infty} \frac{3^n}{3^{3n}} $$ simplifies to $$ \sum_{n=0}^{\infty} \left(\frac{1}{9}\right)^n $$. This is a geometric series with a common ratio of $$ \frac{1}{9} $$. Since $$ \left|\frac{1}{9}\right| < 1 $$, the series converges. The sum of the series is $$ \frac{9}{8} $$.

Given the series answer the following questions.
Answer Attempt 2 out of 2
is
converges since a geometric series will
converge if and only if . The sum of the series is with the common ratio
.

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Given the series answer the following questions. Answer Attempt 2 out of 2 is converges since a geometric series will converge if and only if . The sum of the series is with the common ratio .