Question

Let \sum x_n be a series such that $x_n > 0$ for all $n \in \mathbb{N}$. If $\lim_{n \to \infty} \frac{5^n x_n}{2^n} = 0$, then Select one: a. The series \sum x_n is divergent by nth Term Test. b. The series \sum x_n is divergent by Limit Comparison Test. c. Non of them. d. The series \sum x_n is convergent by Limit Comparison Test. e. The series \sum x_n is convergent by nth Term Test.

          Let \sum x_n be a series such that $x_n > 0$ for all $n \in \mathbb{N}$. If $\lim_{n \to \infty} \frac{5^n x_n}{2^n} = 0$, then
Select one:
a.
The series \sum x_n is divergent by nth Term Test.
b.
The series \sum x_n is divergent by Limit Comparison Test.
c. Non of them.
d.
The series \sum x_n is convergent by Limit Comparison Test.
e.
The series \sum x_n is convergent by nth Term Test.
        
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Let ∑xn be a series such that xn > 0 for all n ∈ℕ. If limn →∞(5^n xn)/(2^n) = 0, then
Select one:
a.
The series ∑xn is divergent by nth Term Test.
b.
The series ∑xn is divergent by Limit Comparison Test.
c. Non of them.
d.
The series ∑xn is convergent by Limit Comparison Test.
e.
The series ∑xn is convergent by nth Term Test.

Added by Jenny H.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Let sum x_(n) be a series such that x_(n)>0 for all ninN. If lim_(n->infty )(5^(n)x_(n))/(2^(n))=0, then Select one: a. The series sum x_(n) is divergent by nth Term Test. b. The series sum x_(n) is divergent by Limit Comparison Test. c. Non of them. d. The series sum x_(n) is convergent by Limit Comparison Test. e. The series sum x_(n) is convergent by nth term Test. Let Z xp be a series such that X,> O for all n E N. If lim n 2n =0,then Select one: Oa. The series >' xn is divergent by nth Term Test. The series , is divergent by Limit Comparison Test. O c. Non of them O d. The series x, is convergent by Limit Comparison Test. e The series , is convergent by nth Term Test.
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Transcript

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00:01 In this question, we are asked to determine the convergence of the given series.
00:08 And to do that, we will use the limit comparison test.
00:12 And we will compare this series to the series.
00:18 Now we are going to choose the highest power of n in the numerator, which is going to be n to the first, and we will choose the highest power of n in the denominator, which is going to be n -cube.
00:36 And we will compare, and this is the number, this series simplifies to the series 1 over n squared and this is a p series and by the p test we know that the series converges.
00:53 The series 1 over n squared converges.
00:57 Now by the limit comparison test we need to calculate the limit of a .n where a .n is a general term of r series over bn where bn is a general term of the other series.
01:16 So in our case this is going to be the limit of 10 n plus 1 over n times n plus 1 times n plus 2 divided by 1 over n squared as n goes to infinity...
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