https://file.aiquickdraw.com/mathbot/file/1728694988272te1mkqts.jpg

Let's solve each series one by one to determine if they are convergent or divergent. If they are convergent, we'll also determine their sums.

### a) \(\sum_{n=1}^{\infty} \frac{(-3)^{n-1}}{4^n}\)

This is a geometric series of the form:

\[
\sum_{n=1}^{\infty} ar^{n-1}
\]
Where \(a = \frac{1}{4}\) and \(r = \frac{-3}{4}\) (since \( (-3)^{n-1} \) implies the ratio involves \(-3/4\)).

#### Convergence test:
A geometric series converges if \(|r| < 1\). Here, \(|r| = \left|\frac{-3}{4}\right| = \frac{3}{4}\), which is less than 1. Thus, this series **converges**.

#### Sum:
The sum of a convergent geometric series is given by:

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

Substitute the values of \(a = \frac{1}{4}\) and \(r = \frac{-3}{4}\):

\[
S = \frac{\frac{1}{4}}{1 - \left(\frac{-3}{4}\right)} = \frac{\frac{1}{4}}{1 + \frac{3}{4}} = \frac{\frac{1}{4}}{\frac{7}{4}} = \frac{1}{7}
\]

Thus, the sum of the series is \( \frac{1}{7} \).

---

### b) \(\sum_{k=0}^{\infty} e^{-2k}\)

This series can be rewritten as:

\[
\sum_{k=0}^{\infty} \left(e^{-2}\right)^k
\]

This is a geometric series with \(a = 1\) (the first term) and \(r = e^{-2}\).

#### Convergence test:
A geometric series converges if \(|r| < 1\). Since \(e^{-2}\) is a positive number less than 1, the series **converges**.

#### Sum:
For a geometric series, the sum is:

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

Substituting \(a = 1\) and \(r = e^{-2}\):

\[
S = \frac{1}{1 - e^{-2}}
\]

Thus, the sum of the series is \( \frac{1}{1 - e^{-2}} \).

---

### c) \(\sum_{n=4}^{\infty} \frac{2^{2n+1}}{5^n}\)

First, let's simplify the general term:

\[
\frac{2^{2n+1}}{5^n} = \frac{2 \cdot 4^n}{5^n} = 2 \left(\frac{4}{5}\right)^n
\]

So the series becomes:

\[
2 \sum_{n=4}^{\infty} \left(\frac{4}{5}\right)^n
\]

This is a geometric series with \(a = \left(\frac{4}{5}\right)^4\) (since the index starts from \(n = 4\)) and the common ratio \(r = \frac{4}{5}\).

#### Convergence test:
A geometric series converges if \(|r| < 1\). Here, \(\left|\frac{4}{5}\right| = \frac{4}{5}\), which is less than 1, so the series **converges**.

#### Sum:
For a geometric series that starts at \(n = N\), the sum is given by:

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

Here, \(a = \left(\frac{4}{5}\right)^4\) and \(r = \frac{4}{5}\), so the sum is:

\[
S = 2 \times \frac{\left(\frac{4}{5}\right)^4}{1 - \frac{4}{5}} = 2 \times \frac{\left(\frac{4}{5}\right)^4}{\frac{1}{5}} = 10 \times \left(\frac{4}{5}\right)^4 = 10 \times \frac{256}{625} = \frac{2560}{625}
\]

Thus, the sum of the series is \( \frac{2560}{625} \).

---

### Summary:
- a) The series converges, and its sum is \( \frac{1}{7} \).
- b) The series converges, and its sum is \( \frac{1}{1 - e^{-2}} \).
- c) The series converges, and its sum is \( \frac{2560}{625} \).

Let me know if you need further clarification!

---

Here are 5 related questions:
1. How do you determine if a geometric series is convergent?
2. Can a geometric series with a common ratio of 1 or greater ever converge?
3. How do you apply the formula for the sum of a geometric series that starts at an index greater than 1?
4. What happens to the sum of a geometric series as the ratio \(r\) approaches 1?
5. Can geometric series with negative common ratios have real sums?

**Tip:** Always check the absolute value of the common ratio in a geometric series to determine convergence.

Determine whether the given geometric series is convergent or divergent. If the series is convergent, determine its sum.

Explore the convergence and sum of three different geometric series. Learn how to determine if a series converges and find its sum using the formula for geometric series. Suitable for advanced high school and college-level math students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation