Convergence or Divergence
Which of the series in Exercis...

Question

Convergence or Divergence

Which of the series in Exercises 57–64 converge, and which diverge? Give reasons for your answers.

∑ (from n = 1 to ∞) [(-1)ⁿ (n!)ⁿ] / [n^(n²)]

Accepted Answer

Determine whether the series sum of N equals 1 to infinity, of -1 to the n, 3N factorial to the n, divided by 3N race the 3n squared converges or diverges. And we have two answers being converges or diverges. Now, to solve this, we're going to make use of the root test. The root test tells us for a series. A subn we compute L, which is the limit, as N goes to infinity, of the nth roots of the series A N. If L is less than 1, the series converges, absolutely. L is greater than 1, the series diverges, or L equals 1 test is inconclusive, and we'd have to do a different test. Now let's let the nth term of our series be A sub N. A subn in our case will be -1 to the n. 3n factorial raised to the N divided by 3n raised to the 3 N squared. Now, we'll take the absolute value in this, since the root test actually uses the absolute value. And we can simplify this by using the nth roots. You have the nth root of a sub n. Now, keep in mind the absolute value we had before is actually going to remove our -1 races to the end, so we can actually take this part of it out. Now we have the nth root of 3 n factorial race to the n divided by 3 n raced to the 3 n squared. From here we can simplify. Since we are taking the nth root, we will raise everything to the 1 divided by n power. Giving us 3 n factorial race the n divided by 3 n race 3 n squared, all race the 1 divided by n power. Now we can simplify by using our exponent rules. This simplifies to give us 3 in factorial. Divided by 3 in race to the 3 in power. From here we can find our limit L. So L is equivalent. 2 Our function here Our expression. Now we can simplify our expression by expanding 3 and factorial is just. 1 multiplied by 2 multiplied by 3, and so on. Until you get 23 in. Divided by 3 n multiplied by 3 n and so on. Now, this is a product of 3 in fractions, where each form becomes K divided by 3 N. We can then write this as the following. 1 divided by 3 n. Multiplied by 2 divided by 3 in, multiplied by 3 divided by 3 in, and so on. Which turns into 1 divided by 3 n. Now this should be greater than 0. Now we can make use of the squeeze theorem to determine the limit. We'll take the limit. Of 1 divided by 3 n as n goes to infinity. Now this limit is 0. By the squeeze theorem we can then conclude that the limit. As N approaches zero of our series, 3 n factorial divided by 3 n raised to the 3 n. is equals to 0. Which means that l Is less than one. Based on what the root test tells us, this series converges absolutely. That means the answer to our problem is answer A. OK, I hope to help you solve the problem. Thank you for watching. Goodbye.

Convergence or DivergenceWhich of the series in Exercises 57–64 converge, and which diverge? Give reasons for your answers.∑ (from n = 1 to ∞) [(-1)ⁿ (n!)ⁿ] / [n^(n²)]

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Convergence or Divergence
Which of the series in Exercises 57–64 converge, and which diverge? Give reasons for your answers.
∑ (from n = 1 to ∞) [(-1)ⁿ (n!)ⁿ] / [n^(n²)]