We need to find the convergence set for the given power series. Let's solve the problems one by one.

1. Problem Description

We need to find the convergence set for the given power series. Let's solve the problems one by one.

2. Solution Steps

Problem 1: $\sum_{n=1}^{\infty} \frac{x^n}{(n-1)!}$

We use the ratio test.

$a_n = \frac{x^n}{(n-1)!}$

$a_{n+1} = \frac{x^{n+1}}{n!}$

$\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| = \lim_{n \to \infty} |\frac{x^{n+1}}{n!} \cdot \frac{(n-1)!}{x^n}| = \lim_{n \to \infty} |\frac{x}{n}| = |x| \lim_{n \to \infty} \frac{1}{n} = |x| \cdot 0 = 0$

Since the limit is 0 for all $x$, the series converges for all $x$.

Problem 2: $\sum_{n=1}^{\infty} \frac{x^n}{3^n}$

$a_n = \frac{x^n}{3^n} = (\frac{x}{3})^n$

We use the ratio test.

$\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| = \lim_{n \to \infty} |\frac{(\frac{x}{3})^{n+1}}{(\frac{x}{3})^n}| = \lim_{n \to \infty} |\frac{x}{3}| = |\frac{x}{3}|$

For convergence, we need $|\frac{x}{3}| < 1$, which means $|x| < 3$, so $-3 < x < 3$.

Now we check the endpoints.

When $x = 3$, we have $\sum_{n=1}^{\infty} \frac{3^n}{3^n} = \sum_{n=1}^{\infty} 1$, which diverges.

When $x = -3$, we have $\sum_{n=1}^{\infty} \frac{(-3)^n}{3^n} = \sum_{n=1}^{\infty} (-1)^n$, which diverges.

So the interval of convergence is $(-3, 3)$.

Problem 3: $\sum_{n=1}^{\infty} \frac{x^n}{n^2}$

$a_n = \frac{x^n}{n^2}$

We use the ratio test.

$\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| = \lim_{n \to \infty} |\frac{x^{n+1}}{(n+1)^2} \cdot \frac{n^2}{x^n}| = \lim_{n \to \infty} |x \frac{n^2}{(n+1)^2}| = |x| \lim_{n \to \infty} (\frac{n}{n+1})^2 = |x| \cdot 1 = |x|$

For convergence, we need $|x| < 1$, which means $-1 < x < 1$.

Now we check the endpoints.

When $x = 1$, we have $\sum_{n=1}^{\infty} \frac{1}{n^2}$, which converges (p-series with $p=2 > 1$).

When $x = -1$, we have $\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$, which converges (alternating series test).

So the interval of convergence is $[-1, 1]$.

Problem 4: $\sum_{n=1}^{\infty} nx^n$

$a_n = nx^n$

We use the ratio test.

$\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| = \lim_{n \to \infty} |\frac{(n+1)x^{n+1}}{nx^n}| = \lim_{n \to \infty} |x \frac{n+1}{n}| = |x| \lim_{n \to \infty} \frac{n+1}{n} = |x| \cdot 1 = |x|$

For convergence, we need $|x| < 1$, which means $-1 < x < 1$.

Now we check the endpoints.

When $x = 1$, we have $\sum_{n=1}^{\infty} n$, which diverges.

When $x = -1$, we have $\sum_{n=1}^{\infty} n(-1)^n$, which diverges.

So the interval of convergence is $(-1, 1)$.

Problem 5: $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^n}{n^2}$

$a_n = (-1)^{n+1} \frac{x^n}{n^2}$

We use the ratio test.

$\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| = \lim_{n \to \infty} |\frac{(-1)^{n+2} x^{n+1}}{(n+1)^2} \cdot \frac{n^2}{(-1)^{n+1} x^n}| = \lim_{n \to \infty} |x \frac{n^2}{(n+1)^2}| = |x| \lim_{n \to \infty} (\frac{n}{n+1})^2 = |x| \cdot 1 = |x|$

For convergence, we need $|x| < 1$, which means $-1 < x < 1$.

Now we check the endpoints.

When $x = 1$, we have $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n^2}$, which converges (alternating series test).

When $x = -1$, we have $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{(-1)^n}{n^2} = \sum_{n=1}^{\infty} (-1)^{2n+1} \frac{1}{n^2} = \sum_{n=1}^{\infty} -\frac{1}{n^2}$, which converges (p-series with $p=2 > 1$).

So the interval of convergence is $[-1, 1]$.

Problem 6: $\sum_{n=1}^{\infty} (-1)^n \frac{x^n}{n}$

$a_n = (-1)^n \frac{x^n}{n}$

We use the ratio test.

$\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| = \lim_{n \to \infty} |\frac{(-1)^{n+1} x^{n+1}}{n+1} \cdot \frac{n}{(-1)^n x^n}| = \lim_{n \to \infty} |x \frac{n}{n+1}| = |x| \lim_{n \to \infty} \frac{n}{n+1} = |x| \cdot 1 = |x|$

For convergence, we need $|x| < 1$, which means $-1 < x < 1$.

Now we check the endpoints.

When $x = 1$, we have $\sum_{n=1}^{\infty} (-1)^n \frac{1}{n}$, which converges (alternating series test).

When $x = -1$, we have $\sum_{n=1}^{\infty} (-1)^n \frac{(-1)^n}{n} = \sum_{n=1}^{\infty} \frac{1}{n}$, which diverges (harmonic series).

So the interval of convergence is $(-1, 1]$.

Problem 7: $\sum_{n=1}^{\infty} (-1)^n \frac{(x-2)^n}{n}$

$a_n = (-1)^n \frac{(x-2)^n}{n}$

We use the ratio test.

$\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| = \lim_{n \to \infty} |\frac{(-1)^{n+1} (x-2)^{n+1}}{n+1} \cdot \frac{n}{(-1)^n (x-2)^n}| = \lim_{n \to \infty} |(x-2) \frac{n}{n+1}| = |x-2| \lim_{n \to \infty} \frac{n}{n+1} = |x-2| \cdot 1 = |x-2|$

For convergence, we need $|x-2| < 1$, which means $-1 < x-2 < 1$, so $1 < x < 3$.

Now we check the endpoints.

When $x = 1$, we have $\sum_{n=1}^{\infty} (-1)^n \frac{(-1)^n}{n} = \sum_{n=1}^{\infty} \frac{1}{n}$, which diverges (harmonic series).

When $x = 3$, we have $\sum_{n=1}^{\infty} (-1)^n \frac{(1)^n}{n} = \sum_{n=1}^{\infty} (-1)^n \frac{1}{n}$, which converges (alternating series test).

So the interval of convergence is $(1, 3]$.

Problem 8: $\sum_{n=1}^{\infty} \frac{(x+1)^n}{n!}$

$a_n = \frac{(x+1)^n}{n!}$

$a_{n+1} = \frac{(x+1)^{n+1}}{(n+1)!}$

We use the ratio test.

$\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| = \lim_{n \to \infty} |\frac{(x+1)^{n+1}}{(n+1)!} \cdot \frac{n!}{(x+1)^n}| = \lim_{n \to \infty} |\frac{x+1}{n+1}| = |x+1| \lim_{n \to \infty} \frac{1}{n+1} = |x+1| \cdot 0 = 0$

Since the limit is 0 for all $x$, the series converges for all $x$.

3. Final Answer

1. $(-\infty, \infty)$

2. $(-3, 3)$

3. $[-1, 1]$

4. $(-1, 1)$

5. $[-1, 1]$

6. $(-1, 1]$

7. $(1, 3]$

8. $(-\infty, \infty)$