Convergence and Divergence
Which of the sequences {aₙ} i...

Question

Convergence and Divergence

Which of the sequences {aₙ} in Exercises 31–100 converge, and which diverge? Find the limit of each convergent sequence.

aₙ = (xⁿ / (2n + 1))^(1/n),x > 0

Accepted Answer

Welcome back everyone. For the sequence a n equals y to the power of n divided by n2 + 1, all raised to the power of 1 divided by n where y is greater than 0. Determine if it converges and find its limit if it does. A converges to 0, b converges to y. C converges to 2 y, and D diverges. So for this problem, let's begin by applying the nth term test for convergence of sequences. We get the limits and approaches infinity of a n, which is going to be limit. As an approaches infinity of. Y to the power of m. Divided by n2 + 1, raise to the power of 1 divided by n. What we can do is distribute the exponent using the properties of exponents, right, to get the limit as n approaches infinity of y because we get y to the power of n multiplied by 1 divided by n. Divided by N2 + 1, a race to the power of 1 divided by n. Now listeners understand that y is a constant, right? So we can basically take out y and evaluate the limit of the denominator. Soy divided by limitsn approaches infinity. Of n2 + 1 Raise the power of 1 divided by n. And now let's evaluate that remaining limit. Limits n approaches infinity of n2 + 1, so the power of 1 divided by n. And what we're going to do here is simply. Try to substitute an approaches infinity to get infinity. Raise the power of 1 divided by infinity. That's going to be 0, right? It is an indeterminate form. So what we can do is simply apply logarithms. Let's suppose that this limit is equal to l, and we're going to evaluate the natural logarithm of L, which is limit as an approaches infinity. Of ln of n2 + 1 raises the power of 1 divided by n which allows us to apply the power rule of logarithms. This is limit as n approaches infinity of ln of n2 + 1. Divided by n because we bring down the exponent which is 1 divided by n, right? And from here we notice that we get an indeterminate form infinity divided by infinity. The denominator immediately approaches infinity. In the numerator, we have a land of an infinite. Infinitely large number. Because we have that indeterminate form, we can apply Lapiel's rule, so let's go ahead and apply it. Limits and approaches infinity of. Now the derivative of the numerator is going to be 1 divided by n2 + 1 multiplied by n2 + 1 derivative using the chain rule, and in the denominator we have the derivative of n which is just 1. So now let's evaluate the remaining derivative, that's going to be 2 N. We get the limit as n approaches infinity of 2 n. Divided by m2 + 1 and in this case again we get infinity divided by infinity as an indeterminate form so we can apply Lapto's rule once again. Which is going to give us a limit as and approaches infinity of the derivative of 2 and S 2. And the derivative of n2 + 1 is 2 n, right? Notice that we can cancel out 2 and 2, which gives us 1 divided by n, and the limit as n approaches infinity of 1 divided by n is 0 because we are going to have a constant divided by an infinitely large number. So if our limit. Or specifically if ln of our limit is equal to 0, then the limit itself l is going to be e to the power of 0, which is 1. So now we have the value of the limit in the denominator, and we basically get y divided by 1, which is y. So the limit as n approaches infinity of a n is y because this is a finite constant, we can conclude that the sequence converges and it converges to that finite constant, meaning the correct answer is option B converges toy. Thank you for watching.

Convergence and DivergenceWhich of the sequences {aₙ} in Exercises 31–100 converge, and which diverge? Find the limit of each convergent sequence.aₙ = (xⁿ / (2n + 1))^(1/n),x > 0

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Convergence and Divergence
Which of the sequences {aₙ} in Exercises 31–100 converge, and which diverge? Find the limit of each convergent sequence.
aₙ = (xⁿ / (2n + 1))^(1/n),x > 0