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• ## Related Books

 0 answersPost by Melinda Berry on April 1, 2015Hello, could you confirm that this is a geometric series? I thought that a geometric series had to have a common ratio. In this case, what is the common ratio? Thank you.

### Harmonic & P Series

• Harmonic series: , always diverges
• P-series: , check P value for convergence

### Harmonic & P Series

Does the series + [1/8] + [1/27] + [1/64] + ... converge?
• Recognize series as a P - series
• 1 + [1/8] + [1/27] + [1/64] + ... = ∑n = 1 [1/(n3)]
Analyse pp = 3 > 1, thus the series converges
Does the series 1 + [1/(3√{2})] + [1/(3√{3})] + [1/(3√{4})] + ... converge?
• Recognize series as a P - series
• 1 + [1/(3√{2})] + [1/(3√{3})] + [1/(3√{4})] + ... = ∑n = 1 [1/(n1/3)]
Analyse pp = [1/3] ≤ 1, thus the series diverges
Does the series 3 + [3/4] + [1/3] + [3/16] + ... diverge?
• Recognize series as a P - series
• 3 + [3/4] + [1/3] + [3/16] + ... = ∑n = 1 [3/(n2)]
• = 3∑n = 1 [1/(n2)]
Analyse pp = 2 > 1, thus the series converges
Does the series ∑n = 1 [1/(n7/8)] converge or diverge?
Analyse pp = [7/8] ≤ 1, thus the series diverges
Does the series ∑n = 1 [1/(n[91/72])] converge or diverge?
Analyse p
p = [91/72]f1, thus the series converges
Does the series ∑n = 1 [1/(√{n3} )] converge or diverge?
• Alter the series with exponent properties
• n = 1 [1/(√{n3} )] = ∑n = 1 [1/(n[3/2])]
Analyse pp = [3/2] > 1, thus the series converges
Does the series ∑n = 1 [1/(6√{n4})] converge or diverge?
• Alter the series with exponent properties
• n = 1 [1/(6√{n4})] = ∑n = 1 [1/(n[4/6])]
Analyse pp = [4/6] ≤ 1, thus the series diverges
Does the series ∑n = 1 [(2n + 2)/(n2 + n)] converge or diverge?
• Simplify series
• n = 1 [(2n + 2)/(n2 + n)] = ∑n = 1 [(2(n + 1))/(n(n + 1))]
• = ∑n = 1 [2/n]
• = 2 ∑n = 1 [1/n]
Analyse pp = 1 ≤ 1, thus the series diverges
Does the series ∑n = 1 [(√n )/n] converge or diverge?
• Alter the series with exponent properties
• n = 1 [(√n )/n] = ∑n = 1 [1/(n[1/2])]
Analyse pp = [1/2] ≤ 1, thus the series diverges
Does the series ∑n = 1 [(√{n2 + 1} )/(√{n4 + n2} )] converge or diverge?
• Alter the series with exponent properties
• n = 1 [(√{n2 + 1} )/(√{n4 + n2} )] = ∑n = 1 [(√{n2 + 1} )/(√{n2(n2 + 1)} )]
• = ∑n = 1 [(√{n2 + 1} )/(√{n2} √{n2 + 1} )]
• = ∑n = 1 [1/n]
Analyse pp = 1 ≤ 1, thus the series diverges
Does the series 36 + 9 + 4 + [36/16] + ... converge or diverge?
• Examine the series by having the denominator expressed as the powers of 2
• 36 + 9 + 4 + [36/16] + ... = [36/(12)] + [36/(22)] + [36/(32)] + [36/(42)] + ...
• = ∑n = 1 [36/(n2)]
• = 36∑n = 1 [1/(n2)]
Analyse p
p = 2 > 1,thus the series converges

*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.