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

@@ -111,5 +111,10 @@ Then if the *series converges* absolutely then the sum converges.
 
 > Does the series $\sum_{n=1}^\infty (\frac{(-1)^n}{n+5}$  conditionally converge, absolutely converge, or diverge?
 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. 
+2. The above series can be compared to $\frac{1}{n}$ with the limit comparison test. $\lim_{n \to \infty}\dfrac{\frac{1}{n}}{\frac{1}{n+5}} = 1 \ne 0, \infty$. By the L.C.T, $\frac{1}{n+5}$ diverges, so the series does not absolutely converge.
+3. Applying the Alternating Series Test,
+	- $a_n > 0$ for all $n$ - all values in the series are greater than zero
+	- $a_n \ge a_{n+1}$ - the series decreases
+	- $\lim_{n \to \infty} a_n = 0$ - the series approaches zero
+All three conditions hold true, therefore we know that $\sum_{n=1}^\infty \frac{(-1)^n}{n+5}$ conditionally converges.