1. Recognize $\frac{1}{n^2+1} < \frac{1}{n^2}$. 2. Know that $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges (p-series with p=2 > 1). 3. Conclude that $\sum_{n=1}^{\infty} \frac{1}{n^2+1}$ converges by the Direct Comparison Test.

1. Choose $b_n = \frac{n}{n^3} = \frac{1}{n^2}$. 2. Evaluate $\lim_{n\to\infty} \frac{a_n}{b_n} = \lim_{n\to\infty} \frac{\frac{n}{n^3-2}}{\frac{1}{n^2}} = 1$. 3. Since the limit is finite and positive, and $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges, conclude that $\sum_{n=1}^{\infty} \frac{n}{n^3-2}$ converges.

1. Focus on the dominant terms: $2n$ in the numerator and $n^2$ in the denominator. 2. Form $b_n = \frac{2n}{n^2} = \frac{2}{n} = \frac{1}{n}$.

1. Compare to $\frac{1}{\sqrt{n}}$. 2. Note that $\frac{1}{\sqrt{n}-1} > \frac{1}{\sqrt{n}}$. 3. Recognize that $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$ diverges (p-series with p=1/2 < 1). 4. Conclude that $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n} - 1}$ diverges by the Direct Comparison Test.

1. Use the fact that $-1 \le \sin(n) \le 1$. 2. Compare to $\sum_{n=1}^{\infty} \frac{1}{n^2}$. 3. Since $\left|\frac{\sin(n)}{n^2}\right| \le \frac{1}{n^2}$ and $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges, conclude that $\sum_{n=1}^{\infty} \frac{\sin(n)}{n^2}$ converges absolutely by the Direct Comparison Test.

1. Notice that $2^n$ grows faster than $n$. 2. Choose $b_n = \frac{1}{2^n}$. 3. Use the Direct Comparison Test since $\frac{1}{2^n + n} < \frac{1}{2^n}$.

1. Recognize that this is not directly comparable to a p-series or geometric series. 2. Consider the Integral Test (not a comparison test, but relevant). 3. Since $\int_2^{\infty} \frac{1}{x\ln(x)} dx$ diverges, conclude that $\sum_{n=2}^{\infty} \frac{1}{n\ln(n)}$ diverges.

1. Compare to $b_n = \frac{3^n}{4^n} = (\frac{3}{4})^n$. 2. Use the Limit Comparison Test. 3. Since $\sum_{n=1}^{\infty} (\frac{3}{4})^n$ converges (geometric series with |r| < 1), conclude that $\sum_{n=1}^{\infty} \frac{3^n}{4^n - 2^n}$ converges.

Direct Comparison Test: when you can easily show $a_n < b_n$ or $a_n > b_n$. Limit Comparison Test: when it's difficult to find a direct inequality, but the limit of the ratio is easy to compute.

Identify a suitable comparison series ($b_n$) with known convergence/divergence behavior.

For series $\sum a_n$ and $\sum b_n$ with $0 \le a_n \le b_n$, if $\sum b_n$ converges, then $\sum a_n$ converges. If $\sum a_n$ diverges, then $\sum b_n$ diverges.

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

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

The geometric series $\sum_{n=0}^{\infty} ar^n$ converges if $|r| < 1$ and diverges if $|r| \ge 1$.

Find a series $\sum b_n$ whose convergence/divergence is known, and show that $0 \le a_n \le b_n$ (for convergence) or $a_n \ge b_n$ (for divergence).

Find a series $\sum b_n$ and compute $\lim_{n\to\infty} \frac{a_n}{b_n}$. If the limit is finite and positive, the series behave alike.

It requires finding a suitable inequality, which can be difficult. It's inconclusive if the inequality goes the wrong way.

The limit must be finite and positive. If the limit is 0 or infinity, the test is inconclusive.

It ensures that the inequalities used in the theorems are valid. If terms are negative, the comparison may not hold.

L'Hopital's Rule can be used to evaluate the limit $\lim_{n\to\infty} \frac{a_n}{b_n}$ when it results in an indeterminate form.

$\sum_{n=1}^{\infty} \frac{1}{n^p}$

$\sum_{n=0}^{\infty} ar^n$

$\sum a_n$ converges.

$\sum b_n$ diverges.

$\lim_{n\to\infty} \frac{a_n}{b_n}$

Both $\sum a_n$ and $\sum b_n$ either converge or diverge.

$ar^{n-1}$

When $p > 1$

When $p \le 1$

When $|r| < 1$