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 = (1/n) ∫₁ⁿ (1/x) dx

Accepted Answer

Welcome back everyone. Determine whether the sequencedn equals 1 divided by an integral from 1 ton of 1 divided by x plus 1 dx. Converges or diverges. If it converges, find its limit. A converges to 1/2. B converges to 1. C converges to 0, and D diverges. So for this problem if we want to determine whether the sequence converges, we are going to apply the nth term test. We want to evaluate the limit as n approaches infinity of the n, which in this case is going to be the limits and approaches infinity. Of 1 divided by n. Integral from 1 to m of 1 divided by x plus 1 dx. Let's go ahead and evaluate the integral. So we have integral from one. Up ton of 1 divided by x plus 1 d x. Let's use the substitution and let's suppose that U is x + 1. Find the derivative of u with respect to x. That's the derivative of x plus 1, which is equal to 1, and this means that DU is equal to dx. Let's change the limits of integration. The lower one is going to be 0 + 1, which is equal to 1, and the upper limit of integration is n + 1. So the integral can be written as the integral from 1 up to n + 1 of duu because the x is DU. Divided by u, and this is an elementary integral. Its value is ln of the absolute value of u, but we know that starting with 1 and going to n + 1, those values are going to be positive. So we get ln of u from 1 up to n + 1. Applying the evaluation theorem, we get Ln of N + 1. Minus ln of 1. Which is going to be equal to LM of M + 1, right? Now from here we know that the integral is ln of m + 1. And what we're going to do is simply substitute the result into our limit. We get limit. As n approaches infinity. Of 1 divided by n multiplied by ln of n + 1. Notice that in the numerator, if we rewrite this as one single fraction, we can basically show that our numerator is ln of N + 1, right? And in the denominator we have N. So in the numerator asn approaches infinity, Ln of infinity is going to be infinity, and in the denominator we will have infinity as well. So this limit, let's call it L. Has an indeterminate form infinity divided by infinity meaning we can apply Lappito's rule and according to Lapito's rule we once differentiate the numerator and the denominator so we will have limit as n approaches infinity of 1 divided by n plus 1 multiplied by n plus 1 derivative divided by n derivative. And this is equal to limit as n approaches infinity of 1 divided by n + 1. The derivative of n plus 1 is 1, and the derivative of n is also 1, right? And now asn approaches infinity, the denominator is infinity plus 1, which is infinity, and the constant divided by infinity gives us a limit of zero. Well done. And now according to the nth term test, if our limit is finite, the sequence converges, and it converges to the value of the limit. So the answer is option C converges to 0. 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 = (1/n) ∫₁ⁿ (1/x) dx

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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 = (1/n) ∫₁ⁿ (1/x) dx