First, take the absolute value of the terms in the series, i.e., consider $\sum |a_n|$.

If $\sum |a_n|$ converges, then $\sum a_n$ is absolutely convergent.

If $\sum |a_n|$ diverges, but $\sum a_n$ converges, then $\sum a_n$ is conditionally convergent.

It's often easier to determine absolute convergence first. If a series is absolutely convergent, you don't need to check for conditional convergence.

The Alternating Series Test can be used to show that an alternating series converges. If the absolute value of that series diverges, then the original series is conditionally convergent.

Comparison tests, like the Direct Comparison Test, can be used to determine if $\sum |a_n|$ converges or diverges, thus helping to establish absolute convergence.

Since $-1 \leq \sin(n) \leq 1$, taking the absolute value means $|sin(n)| \leq 1$. This is useful for comparison tests.

A harmonic series is $\sum_{n=1}^{\infty} \frac{1}{n}$, and it's a classic example of a divergent series. It's often used for comparison.

Absolutely convergent series have better error estimation properties than conditionally convergent series, as rearranging terms in a conditionally convergent series can change its sum.

Take the absolute value of the terms and then apply convergence tests. If the absolute value converges, the series is absolutely convergent.

If $a_n$ is decreasing and $\lim_{n \to \infty} a_n = 0$, then $\sum (-1)^n a_n$ converges.

If $0 \leq a_n \leq b_n$ and $\sum b_n$ converges, then $\sum a_n$ converges. If $a_n \geq b_n \geq 0$ and $\sum b_n$ diverges, then $\sum a_n$ diverges.

If $\lim_{n \to \infty} \frac{a_n}{b_n} = c$, where $0 < c < \infty$, then $\sum a_n$ and $\sum b_n$ either both converge or both diverge.

The series $\sum_{n=1}^{\infty} \frac{1}{n^p}$ converges if $p > 1$ and diverges if $p \leq 1$.

Let $L = \lim_{n \to \infty} |\frac{a_{n+1}}{a_n}|$. If $L < 1$, the series converges absolutely. If $L > 1$, the series diverges. If $L = 1$, the test is inconclusive.

Let $L = \lim_{n \to \infty} \sqrt[n]{|a_n|}$. If $L < 1$, the series converges absolutely. If $L > 1$, the series diverges. If $L = 1$, the test is inconclusive.

If $\sum |a_n|$ converges, then $\sum a_n$ converges.

If $\lim_{n \to \infty} a_n \neq 0$, then the series $\sum a_n$ diverges.

If $f(x)$ is continuous, positive, and decreasing on $[1, \infty)$, then $\sum_{n=1}^{\infty} f(n)$ and $\int_{1}^{\infty} f(x) dx$ either both converge or both diverge.

If a series is absolutely convergent, then any rearrangement of the series converges to the same sum.

1. Take absolute value: $\sum_{n=1}^{\infty} \frac{1}{n^2}$. 2. This is a convergent p-series (p=2). 3. Therefore, the series is absolutely convergent.

1. Take the absolute value: $\sum_{n=1}^{\infty} \frac{|\cos(n)|}{n^2}$. 2. Since $|\cos(n)| \leq 1$, compare to $\sum_{n=1}^{\infty} \frac{1}{n^2}$. 3. The p-series converges, so the original series is absolutely convergent.

1. Verify the series converges using Alternating Series Test. 2. Take the absolute value of the terms. 3. Show the absolute value series diverges. 4. Conclude it's conditionally convergent.

1. Alternating Series Test shows convergence. 2. Absolute value gives $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$, a divergent p-series (p=1/2). 3. Conditionally convergent.

1. Alternating Series Test shows convergence. 2. Absolute value gives $\sum_{n=1}^{\infty} \frac{n}{n^2 + 1}$, which diverges by Limit Comparison Test with $\frac{1}{n}$. 3. Conditionally convergent.

1. Take absolute value: $\sum_{n=1}^{\infty} \frac{|\sin(n)|}{n!}$. 2. Since $|\sin(n)| \leq 1$, compare to $\sum_{n=1}^{\infty} \frac{1}{n!}$. 3. Ratio Test shows $\sum_{n=1}^{\infty} \frac{1}{n!}$ converges. 4. Absolutely convergent.

1. Alternating Series Test shows convergence. 2. Absolute value gives $\sum_{n=2}^{\infty} \frac{1}{\ln(n)}$, which diverges by Comparison Test with $\frac{1}{n}$. 3. Conditionally convergent.

1. Take absolute value: $\sum_{n=1}^{\infty} \frac{n^2}{2^n}$. 2. Apply Ratio Test. 3. The series converges absolutely.

1. Alternating Series Test shows convergence. 2. Absolute value gives $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n} + n}$, which converges by Limit Comparison Test with $\frac{1}{n^{3/2}}$. 3. Absolutely convergent.

1. Alternating Series Test fails since $\lim_{n \to \infty} \frac{n+1}{n} = 1 \neq 0$, so the series diverges. 2. No need to check absolute convergence.