Answer:
∑ (-1)ⁿ⁺³ 1 / (n^½)
∑ (-1)³ⁿ 1 / (8 + n)
Step-by-step explanation:
If ∑ an is convergent and ∑│an│is divergent, then the series is conditionally convergent.
Option A: (-1)²ⁿ is always +1. So an =│an│and both series converge (absolutely convergent).
Option B: bn = 1 / (n^⁹/₈) is a p series with p > 1, so both an and │an│converge (absolutely convergent).
Option C: an = 1 / n³ isn't an alternating series. So an =│an│and both series converge (p series with p > 1). This is absolutely convergent.
Option D: bn = 1 / (n^½) is a p series with p = ½, so this is a diverging series. Since lim(n→∞) bn = 0, and bn is decreasing, then an converges. So this is conditionally convergent.
Option E: (-1)³ⁿ = (-1)²ⁿ (-1)ⁿ = (-1)ⁿ, so this is an alternating series. bn = 1 / (8 + n), which diverges. Since lim(n→∞) bn = 0, and bn is decreasing, then an converges. So this is conditionally convergent.
1a) 8 / (1/2) = 16 * 3 = 48
1b) 3sqrt(49) = 3 * 7 = 21
1c) (5+2)(-8) / (-2)^3 -3
(7*-8) / (-8 -3)
-56/-11
56/11
Easiest ways to solve them are to take the denominator and numerator and and divide the one you are asked to use to answer and divide and see if it works on both.
Example: first 105 / 15 = 7 ) Then: 105 / 7 = 15
119 ) 119 / 7 = 17
Hope I helped. :D
