Let's solve each series step by step to determine whether it converges and, if so, find its sum. Below, I will start solving the questions one by one:

Problem 3

$\sum_{k=1}^\infty \left( -\frac{3}{4} \right)^{k-1}$

Step 1: Recognize the type of series

This is a geometric series with the first term $a = 1$ (when $k = 1$) and common ratio $r = -\frac{3}{4}$.

Step 2: Check for convergence

A geometric series converges if $|r| < 1$. Here, $|r| = \frac{3}{4} < 1$, so the series converges.

Step 3: Sum of a convergent geometric series

The sum of a geometric series is given by: $S = \frac{a}{1 - r}$ Substitute $a = 1$ and $r = -\frac{3}{4}$: $S = \frac{1}{1 - (-\frac{3}{4})} = \frac{1}{1 + \frac{3}{4}} = \frac{1}{\frac{7}{4}} = \frac{4}{7}$

Result for Problem 3: The series converges, and the sum is $\frac{4}{7}$.

Problem 4

$\sum_{k=1}^\infty \left(\frac{2}{3}\right)^{k+2}$

Step 1: Simplify the expression

The term is $\left(\frac{2}{3}\right)^{k+2}$, which can be rewritten as: $\left(\frac{2}{3}\right)^{k+2} = \left(\frac{2}{3}\right)^2 \cdot \left(\frac{2}{3}\right)^k$ So the series becomes: $\sum_{k=1}^\infty \left(\frac{2}{3}\right)^2 \cdot \left(\frac{2}{3}\right)^k$

Step 2: Factor out the constant

The constant $\left(\frac{2}{3}\right)^2 = \frac{4}{9}$ can be factored out: $\sum_{k=1}^\infty \left(\frac{2}{3}\right)^{k+2} = \frac{4}{9} \sum_{k=1}^\infty \left(\frac{2}{3}\right)^k$

Step 3: Recognize the geometric series

The remaining series is a geometric series with $a = \left(\frac{2}{3}\right)$ (the first term when $k = 1$) and common ratio $r = \frac{2}{3}$.

Step 4: Check for convergence

Since $|r| = \frac{2}{3} < 1$, the series converges.

Step 5: Sum of the geometric series

The sum of the geometric series is: $S = \frac{a}{1 - r} = \frac{\frac{2}{3}}{1 - \frac{2}{3}} = \frac{\frac{2}{3}}{\frac{1}{3}} = 2$

Step 6: Multiply by the constant

Multiply the sum by the factor $\frac{4}{9}$: $\text{Total Sum} = \frac{4}{9} \cdot 2 = \frac{8}{9}$

Result for Problem 4: The series converges, and the sum is $\frac{8}{9}$.

I will proceed with solving Problems 5–14 next. Let me know if you'd like me to continue immediately!