From c20ae9b700c8651ad32005d4059eab7a1c65645e Mon Sep 17 00:00:00 2001 From: arc Date: Wed, 1 Oct 2025 12:39:00 -0600 Subject: [PATCH] vault backup: 2025-10-01 12:39:00 --- education/math/MATH1220 (calc II)/Sequences.md | 5 ++++- 1 file changed, 4 insertions(+), 1 deletion(-) diff --git a/education/math/MATH1220 (calc II)/Sequences.md b/education/math/MATH1220 (calc II)/Sequences.md index a8a0075..b3a4bd2 100644 --- a/education/math/MATH1220 (calc II)/Sequences.md +++ b/education/math/MATH1220 (calc II)/Sequences.md @@ -82,4 +82,7 @@ $$ = \sum_{n=1}^\infty 35*(\frac{1}{7})^n*2^{n-1}$$ 2. Pull a $7$ out of the bottom, making use of the fact that $(\frac{1}{7})^n = \frac{1}{7}(\frac{1}{7})^{n-1}$ $$ = \sum_{n=1}^\infty \frac{35}{7}(\frac{1}{7})^{n-1}* 2^{n-1} = \sum_{n = 1}^\infty \frac{35}{7}(\frac{2}{7})^{n-1} $$ 3. This is now of the form $\sum_{n=1}^\infty ar^{n-1}$, so: -$$\sum_{n=1}^{\infty}35(7^{-n} * 2^{n-1}) = \dfrac{\frac{35}{7}}{1}$$ \ No newline at end of file +$$\sum_{n=1}^{\infty}35(7^{-n} * 2^{n-1}) = \dfrac{\frac{35}{7}}{1-\frac{2}{7}}$$ + +# Divergence Test +If $\lim_{n \to \infty} a_n \ne 0$ then $\sum_{a} \ No newline at end of file