From 24c71fa6ac7b5dfa8a5f4ff0195e0726dbbf8997 Mon Sep 17 00:00:00 2001 From: arc Date: Wed, 7 Jan 2026 11:43:08 -0700 Subject: [PATCH] vault backup: 2026-01-07 11:43:08 --- education/math/MATH1220 (calc II)/Sequences and Series.md | 6 +++++- 1 file changed, 5 insertions(+), 1 deletion(-) diff --git a/education/math/MATH1220 (calc II)/Sequences and Series.md b/education/math/MATH1220 (calc II)/Sequences and Series.md index cdcb0b7..0ea5291 100644 --- a/education/math/MATH1220 (calc II)/Sequences and Series.md +++ b/education/math/MATH1220 (calc II)/Sequences and Series.md @@ -150,6 +150,10 @@ $$ |r| < 1 \to |\frac{x}{3}| < 1 \to |x| < 3 $$ This means that the interval of convergence is $(-3, 3)$, given it diverges at both interval endpoints. ## Integral test for series -The integral test determines if an infinite series converges or diverges by comparing it to an improper integral. If the integral diverges ($= \infty$), then the series also diverges. If the integral converges, then the series is +The integral test determines if an infinite series converges or diverges by comparing it to an improper integral. If the integral diverges ($= \infty$), then the series also diverges. If the integral converges, then the series also converges. This is somewhat unique because it proves both convergence and divergence, but has the caveat that the integral must be feasibly solvable. Given the series $\sum_{n=0}^\infty a_n$, the improper integral would be of the below form: +$$ \int_0^\infty a_x dx$$ +Effectively, substitute $n$ for $x$, and swap the summation for an integral. The lower bound of the integral would match the lower bound of the series, and the upper bound will be infinity. + +The integral test only applies if the series is positive, continuous, and decreasing along the entire interval. \ No newline at end of file