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# find the interval of convergence of the series?

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- ted sLv 71 month agoBest Answer
the Ratio Test { try it } yields | x - 2 | < 1 ===> x in ( 1 , 3 )....check the endpoints to see that the series converges there also...thus x in [ 1 , 3 ]

- JOHNLv 71 month ago
u(n) = [(x – 2)^(n – 1)]((n – 1)⁵ + 1]

u(n + 1) = [(x – 2)^n](n⁵ + 1)

|u(n + 1)/u(n)| = |x – 2| x [((n – 1)⁵ + 1]/ (n⁵ + 1)

Lim|u(n + 1)/u(n)| n → ∞ = |x – 2|

Series converges if |x – 2| < 1 ie if – 3 < x < 3

(diverges if |x – 2| > 0

If x = 2 series is 0 + 0 + 0 …=0

So series converges for {2} ∪ {x|– 3 < x < 3}.

- nyphdinmdLv 71 month ago
If x > 2 or x < 1, the numerator become infinitely large (x > 2) as n --> infinite or infinitely negative ( x < 1) as n-->infinity. So the interval of convergence is 1 <= x <=2

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