From edb083d2a8bbdf083ece55cca37d6663bf0d51e0 Mon Sep 17 00:00:00 2001 From: arc Date: Mon, 6 Oct 2025 12:34:06 -0600 Subject: [PATCH] vault backup: 2025-10-06 12:34:06 --- education/math/MATH1220 (calc II)/Sequences.md | 10 ++++++++-- 1 file changed, 8 insertions(+), 2 deletions(-) diff --git a/education/math/MATH1220 (calc II)/Sequences.md b/education/math/MATH1220 (calc II)/Sequences.md index a038d5b..576f129 100644 --- a/education/math/MATH1220 (calc II)/Sequences.md +++ b/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. +