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

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

$\sum a_n$ 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 $|r| < 1$

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.

Direct Comparison: Requires direct inequality ($a_n \le b_n$). Limit Comparison: Compares the limit of the ratio of terms. Direct Comparison: More straightforward when applicable. Limit Comparison: Useful when direct inequality is hard to establish.

p-series: Form $\sum \frac{1}{n^p}$, converges if $p > 1$. Geometric series: Form $\sum ar^n$, converges if $|r| < 1$. Both: Useful for comparison tests. p-series: Depends on exponent p. Geometric series: Depends on common ratio r.

$\sum b_n$ converges: Implies $\sum a_n$ converges. $\sum a_n$ diverges: Implies $\sum b_n$ diverges. The first shows convergence, the second shows divergence.

Finite positive number: Both series behave alike (both converge or both diverge). Zero or infinity: The test is inconclusive and a different comparison series must be found.

$p>1$: $\sum b_n$ converges, useful for showing convergence of a smaller series. $p \le 1$: $\sum b_n$ diverges, useful for showing divergence of a larger series.

Direct Comparison: Compares series to another series. Integral Test: Relates series convergence to integral convergence. Both: Used to determine convergence/divergence. Integral Test: Requires function to be continuous, positive, and decreasing.

r=0.5: Converges because |0.5| < 1. r=2: Diverges because |2| > 1. Convergence depends on the absolute value of r being less than 1.

If $a_n > b_n$ and $\sum a_n$ converges, the test is inconclusive. If $a_n > b_n$ and $\sum b_n$ diverges, then $\sum a_n$ diverges.

Comparison tests are typically used for series with positive terms. Alternating series often require the Alternating Series Test.

Obvious inequality: Application is straightforward. Requires manipulation: More complex, requires algebraic skills to establish the inequality.