EXAMPLE 2For what values of x does the series below converge?(х - 3)"Σ00n=1SOLUTIONLet an =(х - 3)" / п. Thenan(х- 3)"1|x - 3 ||as n → 0By the Ratio test, the given series is absolutely convergent, andtherefore convergent, when | x - 3 | <and divergent when | x - 3 |>Now| x - 3 | < 1 →< x - 3 <The Ratio Test gives no information when | x - 3 | = 1, so we mustconsider x = 2 and x = 4 separately. If we put x = 4 in the series, itbecomes Ethe harmonic series, which diverges. If x = 2,the series is E (-1)" / n, which ( converges v by the AlternatingSeries Test. Thus the given power series converges for 2 [s +v x4.

EXAMPLE 2 For what values of x does the series below converge? 0 (x - 3)" Σ n=1 SOLUTION Let an = (x - 3)" / n. Then %3D an (x - 3)"| 1 | x - 3 || as n → o By the Ratio test, the given series is absolutely convergent, and therefore convergent, when |x - 3 | < and divergent when | x - 3 | > . Now | x - 3 | < 1 → < x - 3 < < x < so the series converges when < x < and diverges when x < or x > The Ratio Test gives no information when | x - 3 | = 1, so we must consider x = 2 and x = 4 separately. If we put x = 4 in the series, it becomes E , the harmonic series, which diverges. If x = 2, the series is E (-1)" / n, which ( converges )- by the Alternating Series Test. Thus the given power series converges for 2 (s +) x < :)v 4.