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how to solve ∑((-5)^-n) n=1 to infinity?do i have it right?
The question asks for absolute convergence, conditional convergence, or diverges.
abs((-5)^-n)=> ∑(1/((-5)^-n)) so it would converge by the geometric series test because r=1/5 <1
So would this be conditionally convergent or absolutely convergent?
1 Answer
- Ron WLv 71 decade ago
Actually, for the given series (which is, as you say, a geometric series) r = -1/5. But the condition for convergence of a geometric sum is |r| < 1, so this series converges. It also converges if you replace -5 by 5; then r = 1/5 and, as before, |r| < 1 so the sum of absolute values converges. Therefore, the series is absolutely convergent.
And you meant ∑(1/((-5)^n)), right? (which is the same as ∑(-1/5)^n).)
