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• ## Related Books

### Power Series Convergence

• Power series:
• Power series can be convergence on an interval of x
• Test for power series convergence:
• Use ratio test to check for absolute convergence
• If not absolutely convergent, solve for x value when it is absolutely convergent
• Check x bounds for convergence by plugging in x to create series

### Power Series Convergence

Find the values of x such that ∑n = 1 [((x − 2)n)/n] converges
• Apply the Ratio Test
• limn →∞ | [(an + 1)/(an)] | = limn →∞ | [((x − 2)n + 1)/(n + 1)]( [n/((x − 2)n)] ) | = limn →∞ | [((x − 2)n)/(n + 1)] |[n/(n + 1)][n/n] = 1
| x − 2 | < 1 − 1 < x − 2 < 11 < x < 3
Find the values of x such that ∑n = 1 [((x + 1)n)/(n2)] converges
• Apply the Ratio Test
• limn →∞ | [(an + 1)/(an)] | = limn →∞ | [((x + 1)n + 1)/((n + 1)2)]( [(n2)/((x + 1)n)] ) | = limn →∞ | [((x + 1)n2)/(( n + 1 )2)] |[(n2)/(( n + 1 )2)][(n2)/(n2)] = 1
| x + 1 | < 1 − 1 < x + 1 < 1 − 2 < x < 0
Find a value of x such that ∑n = 1 [((3x2 − 1)n)/(n3/2)] converges
• Apply the Ratio Test
• limn →∞ | [(an + 1)/(an)] | = limn →∞ | [((3x2 − 1)n + 1)/((n + 1)3/2)]( [(n3/2)/((3x2 − 1)n)] ) | = limn →∞ | [((3x2 − 1)n3/2)/(( n + 1 )3/2)] |[(n3/2)/(( n + 1 )3/2)][(n3/2)/(n3/2)] = 1
• Find the domain of convergence
• | 3x2 − 1 | < 1
0 < 3x2< 20 < x2< [2/3]0 < x <√{[2/3]}
Find the values of x such that ∑n = 1 [n/((x − 1)n)] converges
• Apply the Ratio Test
• limn →∞ | [(an + 1)/(an)] | = limn →∞ | [(n + 1)/((x − 1)n + 1)]( [((x − 1)n)/n] ) | = limn →∞ | [(n + 1)/((x − 1)n)] |[(n + 1)/n][n/n] = 1
• Find the domain of convergence
• [1/(| x − 1 |)] < 1
• [1/(x − 1)] < 1x > 2
[1/(x − 1)] >− 1x < 0
Determine if the series ∑n = 1 [(xn)/(en)] converges absolutely, and if so what's the domain of x
• Apply the Ratio Test
• limn →∞ | [(an + 1)/(an)] | = limn →∞ | [((x)n + 1)/(en + 1)]( [(en)/(xn)] ) | = limn →∞ | [x/e] |[x/e]
| [x/e] | < 1 − 1 < [x/e] < 1 − e < x < e
Determine if the series ∑n = 1 [((x + 1)n)/n!] converges absolutely
Apply the Ratio Test
limn →∞ | [(an + 1)/(an)] | = limn →∞ | [((x + 1)n + 1)/((n + 1)!)]( [n!/((x + 1)n)] ) | = limn →∞ | [(x + 1)/(( n + 1 ))] | = 0 < 1 The series converges, but does not converge absolutely
Find a value of x such that ∑n = 1 [(( − 1)n)/(n + 3)] converges
• Apply the Ratio Test
• limn →∞ | [(an + 1)/(an)] | = limn →∞ | [(xn + 1)/(n + 4)]( [(n + 3)/(xn)] ) | = limn →∞ | x[(n + 3)/(n + 4)] ||x|< 1
• Find the domain of convergence
• | x | < 1
• − 1 < x < 1
Test values
Note that if x = − 1, the series converges via Alternate Series Test. Thus the range is actually
− 1 ≤ x < 1
Find the values of x such that ∑n = 1 [(en)/((x − 12)n)] converges
• Apply the Ratio Test
• limn →∞ | [(an + 1)/(an)] | = limn →∞ | [(en + 1)/((x − 12)n + 1)]( [((x − 12)n)/(en)] ) | = limn →∞ | [e/((x − 12))] | < 1| [e/((x − 12))] | < 1
• Find the domain of convergence
• | [e/((x − 12))] | < 1
• [e/((x − 12))] < 1e + 12 < x
[e/((x − 12))] >− 112 − e > x
Determine if the series ∑n = 1 [(xn)/n!] converges absolutely, and if so what's the domain of x
Apply the Ratio Test
limn →∞ | [(an + 1)/(an)] | = limn →∞ | [((x)n + 1)/((n + 1)!)]( [n!/(xn)] ) | = limn →∞ | [x/(n + 1)] |0 < 1 The series converges, but does not converge absolutely
Determine if the series ∑n = 1 [(( − 1)nxn)/(n2)] converges absolutely if so what's the domain of x
• Apply the Ratio Test
• limn →∞ | [(an + 1)/(an)] | = limn →∞ | [(xn + 1)/((n + 1)2)]( [(n2)/(xn)] ) | = limn → ∞ | [(xn2)/(( n + 1 )2)] || x | < 1
• Determine convergence
• | x | < 1
− 1 < x < 1

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.