In Exercises 57–82, use any method to determine whether the se...

Question

In Exercises 57–82, use any method to determine whether the series converges or diverges. Give reasons for your answer.

∑ (from n = 0 to ∞) [((n + 1) / (n + 2))ⁿ]

Accepted Answer

Welcome back everyone. Determine whether the series sigma from n equals 0 to infinity of n + 3 divided by n + 5 to the power of n converges or diverges. A says converges and b says diverges. So for this problem we're going to use the nth term test for divergence. Specifically, what we want to do is evaluate the limit asn approaches infinity of a n. And in this case, a n is the term n + 3 divided by. N + 5 to the power of n. According to the test, if the limit is not equal to 0, the series diverges, otherwise it's inconclusive. So let's go ahead and evaluate this limit. What we are going to do is Separate the whole part so that be a limit as m approaches infinity. We can write the numerator as n + 5 minus 2, right? That's m + 3 divided by n + 5. Which essentially allows us to perform division or separate the whole part so that limit isn approaches infinity of 1 minus 2 divided by N + 5 to the power of n. And now Let's remember one of the limits, specifically limit. As M Approaches infinity of one. Less. X divided by m to the power of m is equal to e to the power of x. So in this case we're going to use substitution. Let's suppose that m is equal to n + 5, right? So if M approaches infinity, then M approaches infinity as well. And now we are going to solve for M, and our exponent is going to be M. -5. So let's call our limit L and L becomes limit. As m approaches infinity of 1 minus 2 divided by m. A raise to the power of m minus 5. Now what we're going to do is use the properties of exponents, so that's a limit. As m approaches infinity of 1 minus 2 divided by m to the power of m. Multiplied by 1 minus 2 divided by m to the power of negative. 5 So now what we can do is separate. This expression into two limits. One of them is going to be. E raise the power of -2 using our limit, so we can write E to the power of -2 multiplied by. And for the other we have limits. M approaches infinity of 1 minus 2 divided by m to the power of -5. Now as m approaches infinity, the fractional term disappears. 2 divided by an infinitely large number is 0, and 1 to the power of -5 is 1. So we get e to the power of -2, which is 1 divided by e squad. And this is not equal to 0, meaning the series diverges, and the answer to this problem is B. Thy virtues. That's our final answer, and thank you for watching.

In Exercises 57–82, use any method to determine whether the series converges or diverges. Give reasons for your answer.∑ (from n = 0 to ∞) [((n + 1) / (n + 2))ⁿ]

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In Exercises 57–82, use any method to determine whether the series converges or diverges. Give reasons for your answer.
∑ (from n = 0 to ∞) [((n + 1) / (n + 2))ⁿ]