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.