vault backup: 2025-10-06 12:34:06

education/math/MATH1220 (calc II)/Sequences.md

@@ -42,7 +42,7 @@ Vocabulary: A **series** is another name for a sum of numbers.
 
 ## The Limit Test
 You can break a series into *partial sums*:
-$$\sum_{n=1}^\infty a_n = a_1 + 1_2 + a_3 + ...$$
+$$\sum_{n=1}^\infty a_n = a_1 + a_2 + a_3 + ...$$
 Given the above series, we can define the following:
 - $S_1 = a_1 = \sum_{i=1}^\infty a_i$
 - $S_2 = a_1 + a_2 = \sum_{i=1}^2 a_i$ 
@@ -105,5 +105,11 @@ A sequence $a_n$ *converges absolutely* if $\sum |a_n|$ converges
 Then if the *series converges* absolutely then the sum converges.
 
 ## Examples
+> Does the series $\sum_{n=1}^\infty (\frac{(-1)^n}{n^2}$  conditionally converge, absolutely converge, or diverge?
+1. $\sum_{n=1}^\infty|\frac{(-1)^n}{n^2}| = \sum_{n=1}^\infty \frac{1}{n^2}$ 
+2. We know that $\sum_{n=1}^\infty \frac{1}{n^2}$ converges by the p test, so the series absolutely converges.
+
 > Does the series $\sum_{n=1}^\infty (\frac{(-1)^n}{n+5}$  conditionally converge, absolutely converge, or diverge?
-1. $\sum_{n=1}^\infty|\frac{(-1)^n}{n+5}| = \sum_{n=1}^\infty \frac{1}{n+5}$ 
+1. Consider $\sum_{n=1}^\infty|\frac{(-1)^n}{n+5}| = \sum_{n=1}^\infty \frac{1}{n+5}$.
+2. The above series can be compared to $\frac{1}{n}$ with the limit comparison test. 
+