Question

The Root and Ratio Test. We want to determine whether the series $\sum_{n=1}^{\infty} (-1)^n \left( 1 - \frac{1}{n} \right)^{n^2}$ converges or diverges. Since $a_n = (-1)^n \left( 1 - \frac{1}{n} \right)^{n^2}$ contains a exponent of $n$, we have decided to apply the Root Test and find that, $\rho = \lim_{n \to \infty} \sqrt[n]{|a_n|} = \lim_{n \to \infty}$ Note: type a simplified ratio in terms of $n$, then evaluate the limit above. Since $\rho$ choose one $1$ we conclude that choose one Note: You can earn partial credit on this problem.

          The Root and Ratio Test.
We want to determine whether the series $\sum_{n=1}^{\infty} (-1)^n \left( 1 - \frac{1}{n} \right)^{n^2}$ converges or diverges.
Since $a_n = (-1)^n \left( 1 - \frac{1}{n} \right)^{n^2}$ contains a exponent of $n$, we have decided to apply the Root Test and find that,
$\rho = \lim_{n \to \infty} \sqrt[n]{|a_n|} = \lim_{n \to \infty}$ 
Note: type a simplified ratio in terms of $n$, then evaluate the limit above.
Since $\rho$ choose one $1$ we conclude that choose one
Note: You can earn partial credit on this problem.
        
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The Root and Ratio Test.
We want to determine whether the series ∑n=1^∞ (-1)^n ( 1 - (1)/(n))^n^2 converges or diverges.
Since an = (-1)^n ( 1 - (1)/(n))^n^2 contains a exponent of n, we have decided to apply the Root Test and find that,
ρ = limn →∞√(|an|) = limn →∞ 
Note: type a simplified ratio in terms of n, then evaluate the limit above.
Since ρ choose one 1 we conclude that choose one
Note: You can earn partial credit on this problem.

Added by Leslie C.

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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The Root and Ratio Test. Converges or diverges Since an = -11- contains an exponent of n, we have decided to apply the Root Test and find that, p = lim(n→∞) √(an) Note: Type a simplified ratio in terms of n, then evaluate the limit above. Since p = 1, we conclude that... Note: You can earn partial credit on this problem.
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Transcript

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00:01 Here in this question we need to determine whether sigma n equal to 1 to infinite n divided by 2 to the power n converges or diverges.
00:09 So first of all we need to apply here l is equal to limit n tends to infinite mod of a n plus 1 divided by a n.
00:21 Where what is a n? a n is nothing but a n is equal to n divided by 2 to the power n.
00:28 Now we can put here so l is equal to limit n tends to infinite mod of a n plus 1 it means basically n plus 1 divided by 2 to the power n plus 1 times 1 divided by a n it means 2 to the power n divided by n ok...
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