The problem asks to determine whether the given series converges or diverges. If a series converges, we need to find its sum. We will examine problems 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, and 14.

1. Problem Description

The problem asks to determine whether the given series converges or diverges. If a series converges, we need to find its sum. We will examine problems 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, and

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

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 $|r| = 4 > 1$, the series diverges.

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

The first series has $a = 1$ and $r = \frac{1}{4}$. The sum is $\frac{1}{1-\frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3}$.

The second series has $a = 1$ and $r = -\frac{1}{5}$. The sum is $\frac{1}{1-(-\frac{1}{5})} = \frac{1}{\frac{6}{5}} = \frac{5}{6}$.

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

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

The second series has $a = (\frac{1}{7})^2 = \frac{1}{49}$ and $r = \frac{1}{7}$. The sum is $\frac{\frac{1}{49}}{1-\frac{1}{7}} = \frac{\frac{1}{49}}{\frac{6}{7}} = \frac{1}{49} \cdot \frac{7}{6} = \frac{1}{42}$.

So the overall sum is $5(1) - 3(\frac{1}{42}) = 5 - \frac{3}{42} = 5 - \frac{1}{14} = \frac{70}{14} - \frac{1}{14} = \frac{69}{14}$.

5. $\sum_{k=1}^{\infty} \frac{k-5}{k+2}$. We can use the divergence test. $\lim_{k\to\infty} \frac{k-5}{k+2} = \lim_{k\to\infty} \frac{1 - \frac{5}{k}}{1 + \frac{2}{k}} = \frac{1}{1} = 1 \neq 0$. 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 $|r| = \frac{9}{8} > 1$, the series diverges.

7. $\sum_{k=2}^{\infty} (\frac{1}{k} - \frac{1}{k-1})$. This is a telescoping series. Let's write out a few terms: $(\frac{1}{2} - \frac{1}{1}) + (\frac{1}{3} - \frac{1}{2}) + (\frac{1}{4} - \frac{1}{3}) + ... = -1 + (\frac{1}{2} - \frac{1}{2}) + (\frac{1}{3} - \frac{1}{3}) + ...$. So the sum is -

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8. $\sum_{k=1}^{\infty} \frac{3}{k} = 3\sum_{k=1}^{\infty} \frac{1}{k}$. This is a harmonic series multiplied by a constant. The harmonic series $\sum_{k=1}^{\infty} \frac{1}{k}$ diverges. Thus, the series diverges.

9. $\sum_{k=1}^{\infty} \frac{k!}{100^k}$. We can use the ratio test. Let $a_k = \frac{k!}{100^k}$. Then $\frac{a_{k+1}}{a_k} = \frac{\frac{(k+1)!}{100^{k+1}}}{\frac{k!}{100^k}} = \frac{(k+1)!}{k!} \cdot \frac{100^k}{100^{k+1}} = (k+1) \cdot \frac{1}{100} = \frac{k+1}{100}$. $\lim_{k\to\infty} \frac{k+1}{100} = \infty > 1$. Thus, the series diverges.

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0. $\sum_{k=1}^{\infty} \frac{2}{(k+2)k}$. We can use partial fraction decomposition: $\frac{2}{k(k+2)} = \frac{A}{k} + \frac{B}{k+2}$. $2 = A(k+2) + Bk$. Let $k = 0$, then $2 = 2A$, so $A = 1$. Let $k = -2$, then $2 = -2B$, so $B = -1$. Thus, $\frac{2}{k(k+2)} = \frac{1}{k} - \frac{1}{k+2}$.

So the series is $\sum_{k=1}^{\infty} (\frac{1}{k} - \frac{1}{k+2}) = (\frac{1}{1} - \frac{1}{3}) + (\frac{1}{2} - \frac{1}{4}) + (\frac{1}{3} - \frac{1}{5}) + (\frac{1}{4} - \frac{1}{6}) + ... = 1 + \frac{1}{2} = \frac{3}{2}$.

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1. $\sum_{k=1}^{\infty} (\frac{e}{\pi})^{k+1}$. This is a geometric series with $a = (\frac{e}{\pi})^2$ and $r = \frac{e}{\pi}$. Since $e \approx 2.718$ and $\pi \approx 3.141$, we have $\frac{e}{\pi} < 1$. Thus, $|r| < 1$ and the series converges. The sum is $\frac{(\frac{e}{\pi})^2}{1-\frac{e}{\pi}} = \frac{\frac{e^2}{\pi^2}}{\frac{\pi - e}{\pi}} = \frac{e^2}{\pi^2} \cdot \frac{\pi}{\pi - e} = \frac{e^2}{\pi(\pi-e)}$.

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2. $\sum_{k=1}^{\infty} \frac{4^{k+1}}{7^{k-1}} = \sum_{k=1}^{\infty} \frac{4^2 \cdot 4^{k-1}}{7^{k-1}} = 16\sum_{k=1}^{\infty} (\frac{4}{7})^{k-1} = 16\sum_{k=0}^{\infty} (\frac{4}{7})^k$. This is a geometric series with $a = 1$ and $r = \frac{4}{7}$. Since $|r| < 1$, the series converges. The sum is $16 \cdot \frac{1}{1-\frac{4}{7}} = 16 \cdot \frac{1}{\frac{3}{7}} = 16 \cdot \frac{7}{3} = \frac{112}{3}$.

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3. $\sum_{k=2}^{\infty} (\frac{3}{(k-1)^2} - \frac{3}{k^2}) = 3\sum_{k=2}^{\infty} (\frac{1}{(k-1)^2} - \frac{1}{k^2})$. This is a telescoping series. Let $j = k - 1$. $3\sum_{k=2}^{\infty} (\frac{1}{(k-1)^2} - \frac{1}{k^2}) = 3[(\frac{1}{1^2} - \frac{1}{2^2}) + (\frac{1}{2^2} - \frac{1}{3^2}) + (\frac{1}{3^2} - \frac{1}{4^2}) + ...] = 3(1) = 3$.

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4. $\sum_{k=6}^{\infty} \frac{2}{k-5}$. Let $j = k - 5$. Then $k = j+5$, and when $k=6$, $j = 1$. So $\sum_{k=6}^{\infty} \frac{2}{k-5} = \sum_{j=1}^{\infty} \frac{2}{j} = 2\sum_{j=1}^{\infty} \frac{1}{j}$. This is a harmonic series multiplied by a constant. The harmonic series diverges. Thus, the series 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

9. Diverges

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0. Converges, Sum = 3/2

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1. Converges, Sum = $e^2/(\pi(\pi-e))$

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2. Converges, Sum = 112/3

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3. Converges, Sum = 3

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4. Diverges