We need to determine whether the given series converge or diverge. If they converge, we need to find their sum. We are given eight series. 1. $\sum_{k=1}^{\infty} (\frac{1}{7})^k$

1. Problem Description

We need to determine whether the given series converge or diverge. If they converge, we need to find their sum. We are given eight series.

1. $\sum_{k=1}^{\infty} (\frac{1}{7})^k$

2. $\sum_{k=1}^{\infty} (-\frac{1}{4})^{-k-2}$

3. $\sum_{k=0}^{\infty} [2(\frac{1}{4})^k + 3(-\frac{1}{5})^k]$

4. $\sum_{k=1}^{\infty} [5(\frac{1}{2})^k - 3(\frac{1}{7})^{k+1}]$

5. $\sum_{k=1}^{\infty} \frac{k-5}{k+2}$

6. $\sum_{k=1}^{\infty} (\frac{9}{8})^k$

7. $\sum_{k=2}^{\infty} (\frac{1}{k} - \frac{1}{k-1})$

8. $\sum_{k=1}^{\infty} \frac{3}{k}$

2. Solution Steps

1. $\sum_{k=1}^{\infty} (\frac{1}{7})^k$

This is a geometric series with $a = \frac{1}{7}$ and $r = \frac{1}{7}$. Since $|r| = |\frac{1}{7}| < 1$, the series converges. The sum of the geometric series is given by

$S = \frac{a}{1-r} = \frac{\frac{1}{7}}{1-\frac{1}{7}} = \frac{\frac{1}{7}}{\frac{6}{7}} = \frac{1}{6}$.

2. $\sum_{k=1}^{\infty} (-\frac{1}{4})^{-k-2} = \sum_{k=1}^{\infty} (-4)^{k+2} = \sum_{k=1}^{\infty} (-4)^2 (-4)^k = 16 \sum_{k=1}^{\infty} (-4)^k$

This is a geometric series with $a = -4$ and $r = -4$. Since $|r| = |-4| = 4 > 1$, the series diverges.

3. $\sum_{k=0}^{\infty} [2(\frac{1}{4})^k + 3(-\frac{1}{5})^k] = 2 \sum_{k=0}^{\infty} (\frac{1}{4})^k + 3 \sum_{k=0}^{\infty} (-\frac{1}{5})^k$

The first series is a geometric series with $a = 1$ and $r = \frac{1}{4}$. Since $|\frac{1}{4}| < 1$, the series converges to $\frac{1}{1-\frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3}$.

The second series is a geometric series with $a = 1$ and $r = -\frac{1}{5}$. Since $|-\frac{1}{5}| < 1$, the series converges to $\frac{1}{1-(-\frac{1}{5})} = \frac{1}{1+\frac{1}{5}} = \frac{1}{\frac{6}{5}} = \frac{5}{6}$.

Thus, the sum is $2 (\frac{4}{3}) + 3 (\frac{5}{6}) = \frac{8}{3} + \frac{15}{6} = \frac{16}{6} + \frac{15}{6} = \frac{31}{6}$.

4. $\sum_{k=1}^{\infty} [5(\frac{1}{2})^k - 3(\frac{1}{7})^{k+1}] = 5 \sum_{k=1}^{\infty} (\frac{1}{2})^k - 3 \sum_{k=1}^{\infty} (\frac{1}{7})^{k+1} = 5 \sum_{k=1}^{\infty} (\frac{1}{2})^k - \frac{3}{7} \sum_{k=1}^{\infty} (\frac{1}{7})^k$

The first series is a geometric series with $a = \frac{1}{2}$ and $r = \frac{1}{2}$. Its sum is $\frac{\frac{1}{2}}{1-\frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1$.

The second series is a geometric series with $a = \frac{1}{7}$ and $r = \frac{1}{7}$. Its sum is $\frac{\frac{1}{7}}{1-\frac{1}{7}} = \frac{\frac{1}{7}}{\frac{6}{7}} = \frac{1}{6}$.

Thus, the sum is $5(1) - \frac{3}{7} (\frac{1}{6}) = 5 - \frac{3}{42} = 5 - \frac{1}{14} = \frac{70-1}{14} = \frac{69}{14}$.

5. $\sum_{k=1}^{\infty} \frac{k-5}{k+2}$

We can apply the divergence test. We have $\lim_{k \to \infty} \frac{k-5}{k+2} = \lim_{k \to \infty} \frac{1 - \frac{5}{k}}{1 + \frac{2}{k}} = \frac{1}{1} = 1 \ne 0$.

Since the limit is not 0, the series diverges.

6. $\sum_{k=1}^{\infty} (\frac{9}{8})^k$

This is a geometric series with $a = \frac{9}{8}$ and $r = \frac{9}{8}$. Since $|\frac{9}{8}| > 1$, the series diverges.

7. $\sum_{k=2}^{\infty} (\frac{1}{k} - \frac{1}{k-1})$

This is a telescoping series. Let's consider the partial sum:

$S_n = \sum_{k=2}^{n} (\frac{1}{k} - \frac{1}{k-1}) = (\frac{1}{2} - \frac{1}{1}) + (\frac{1}{3} - \frac{1}{2}) + (\frac{1}{4} - \frac{1}{3}) + \cdots + (\frac{1}{n} - \frac{1}{n-1})$

$= \frac{1}{n} - 1$.

Therefore, $\lim_{n \to \infty} S_n = \lim_{n \to \infty} (\frac{1}{n} - 1) = 0 - 1 = -1$. The series converges to -

1.

8. $\sum_{k=1}^{\infty} \frac{3}{k} = 3 \sum_{k=1}^{\infty} \frac{1}{k}$

This is a constant multiple of the harmonic series $\sum_{k=1}^{\infty} \frac{1}{k}$, which is known to diverge. Therefore, this series also diverges.

3. Final Answer

1. Converges, Sum = 1/6

2. Diverges

3. Converges, Sum = 31/6

4. Converges, Sum = 69/14

5. Diverges

6. Diverges

7. Converges, Sum = -1

8. Diverges