MAT-232 WA#5 - Name Eleonor Lopez College ID 0515654 Thomas...

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Name: Eleonor Lopez College ID: 0515654 Thomas Edison State College Calculus II (MAT-232) Section no.: 8.6-8.8 Semester and year: Jun 2016 section 8.6 solution 4. wehavethe series k = 0 k 2 k x k letu k = k 2 k x k thenby Ratio test lim k →∞ | u k + 1 u k | = lim k → ∞ | k + 1 2 k + 1 x k + 1 × 2 k k x k | ¿ lim k →∞ | k + 1 2 k x | = | x 2 | Henceby Ratio test the seriesis convergeif | x 2 | < 1 = ¿ | x | < 2 thus radiusof convergenceis R = 2 ¿¿ find convergenceof interval here 2 < x < 2 also wewill ˇ at x =− 2 2 for x =− 2 the seriesbecomes k = 0 ( 1 ) k k this seriesisdivergent also x = 2 the seriesbecomes k = 0 k this seriesisdivergent
hence convergenceof intervalis (− 2,2 ) 10. wehavethe series k = 4 1 k 2 ( 3 x + 2 ) k x + 2 3 ¿ ¿ ( 3 x + 2 ) k + 1 ( k + 1 ) 2 × k 2 ¿ ¿ letu k = 1 k 2 ( 3 x + 2 ) k ,thenby Ratio test lim k →∞ | u k + 1 u k | = lim k → ∞ ¿ ¿ lim k →∞ | ( 3 x + 2 ) k 2 ( k + 1 ) 2 | = | 3 x + 2 | Henceby Ratio test the seriesisconvergeif | 3 x + 2 | < 1 = ¿ | x + 2 3 | < 1 3 thus radiusof convergenceis R = 1 3 ¿¿ find convergenceof interval here 1 3 < x + 2 3 < 1 3 = ¿ 1 < x < 1 3 also wewill ˇ at x = 1 ∧− 1 3 for x =− 1 the seriesbecomes k = 4 ( 1 ) k k 2 this seriesisconvergent also x = 1 3 the series becomes k = 4 1 k 2 this seriesis convergent
hence convergenceof intervalis [− 1, 1 3 ] 12. wehavethe series k = 1 ( 1 ) k k ( 3 x 1 ) k x 1 3 ¿ ¿ ( 3 x 1 ) k + 1 k + 1 × k ¿ ¿ letu k = ( 1 ) k k ( 3 x 1 ) k ,then by Ratiotest lim k→ ∞ | u k + 1 u k | = lim k →∞ ¿ ¿ lim k →∞ | ( 3 x 1 ) k k + 1 | = | 3 x 1 | Henceby Ratio test the seriesis convergeif | 3 x 1 | < 1 = ¿ | x 1 3 | < 1 3 thus radiusof convergenceis R = 1 3 ¿¿ find convergenceof interval here 1 3 < x 1 3 < 1 3 = ¿ 0 < x < 2 3 also wewill ˇ at x = 0 2 3 for x = 0 the seriesbecomes k = 1 1 k this seriesisdivergent also x = 2 3 the seriesbecomes k = 1 ( 1 ) k k thisseriesisconvergent
hence convergenceof intervalis ¿ 16. we havethe series k = 2 ( k ! ) 2 ( 2 k ) ! x 2 k + 1 letu k = ( k ! ) 2 ( 2 k ) ! x 2 k + 1 thenby Ratio test lim k → ∞ | u k + 1 u k | = lim k→ ∞ | ( ( k + 1 ) ! ) 2 ( 2 k + 2 ) ! x 2 k + 3 × ( 2 k ) ! ( k! ) 2 x 2 k + 1 | ¿ lim k →∞ | ( k + 1 ) 2 ( 2 k + 2 )( 2 k + 1 ) x 2 | = | x 2 4 | Henceby Ratio test the seriesisconvergeif | x 2 4 | < 1 = ¿ | x | < 2 thus radiusof convergenceis R = 2 ¿¿ find convergenceof interval here 2 < x < 2 also wewill ˇ at x =− 2 2 for x =− 2 the seriesbecomes k = 0 ( k ! ) 2 ( 2 k ) ! ( 2 ) 2 k + 1 ¿ k = 0 ( k ! ) 2 ( 2 k ) ! 2 2 k + 1 letu k = ( k! ) 2 ( 2 k ) ! 2 2 k + 1 thenby Raabe ' stest u k u k + 1 k ( ¿ 1 ) ¿ = lim k →∞ [ k (− ( k ! ) 2 ( 2 k ) ! 2 2 k + 1 × ( 2 k + 2 ) ! ( ( k + 1 ) ! ) 2 2 2 k + 3 1 )] lim k →∞ ¿
¿ lim [ ¿ k →∞k ( ( 2 k + 2 )( 2 k + 1 ) 4 ( k + 1 ) 2 1 )]= lim k →∞ 2 k 2 2 k 4 ( k + 1 ) 2 =− 1 / 2 < 1 thisseries isdivergent also x = 2 the seriesbecomes k = 0 ( k ! ) 2 ( 2 k ) ! 2 2 k + 1 thisseriesis also same as previous one hence divergent hence convergenceof intervalis (− 2,2 ) 24. wehavethe series k = 0 3 x k 4 k letu k = 3 x k 4 k thenby Ratio test lim k → ∞ | u k + 1 u k | = lim k→ ∞ | 3 x k + 1 4 k + 1 × 4 k 3 x k | ¿ lim k →∞ | x 4 | = | x 4 | Henceby Ratio test the seriesisconvergeif | x 4 | < 1 = ¿ | x | < 4

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