1. The problem is to understand the reasoning behind the Limit Comparison Test (LCT) or the Direct Comparison Test (DCT) for series convergence. 2. The Limit Comparison Test states: Given two series $\sum a_n$ and $\sum b_n$ with positive terms, if the limit $\lim_{n \to \infty} \frac{a_n}{b_n} = c$ where $c$ is a finite number and $c > 0$, then either both series converge or both diverge. 3. This test is useful because it allows us to compare a complicated series $a_n$ to a simpler series $b_n$ whose convergence behavior is known. 4. The Direct Comparison Test states: If $0 \leq a_n \leq b_n$ for all $n$ beyond some index, and if $\sum b_n$ converges, then $\sum a_n$ also converges. Conversely, if $\sum a_n$ diverges and $a_n \leq b_n$, then $\sum b_n$ diverges. 5. Both tests rely on comparing terms of series to known benchmark series to determine convergence or divergence. 6. For example, if $a_n = \frac{1}{n^2 + 1}$ and $b_n = \frac{1}{n^2}$, then $\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{1/(n^2 + 1)}{1/n^2} = \lim_{n \to \infty} \frac{n^2}{n^2 + 1} = 1$, which is finite and positive. 7. Since $\sum b_n = \sum \frac{1}{n^2}$ converges (p-series with $p=2 > 1$), by LCT, $\sum a_n$ also converges. 8. This reasoning shows how LCT helps us conclude convergence by comparing to a known series. Final answer: The Limit Comparison Test and Direct Comparison Test allow us to determine convergence of a series by comparing it to another series with known behavior, using limits or inequalities respectively.