Using the Ratio Test
In Exercises 1–8, use the Ratio Tes...

Question

Using the Ratio Test

In Exercises 1–8, use the Ratio Test to determine whether each series converges absolutely or diverges.

∑(from n=2 to ∞) [(3ⁿ⁺²) / ln(n)]

Accepted Answer

Welcome back everyone. Use the ratio test to determine whether the series sigma from n equals 2 up to infinity of 2 to the power of n + 1 divided by lnn converges absolutely or diverges. A says converges absolutely and b says diverges. So for this problem, let's remember the ratio test. We want to evaluate the limit l, where l is limits and approaches infinity. Of the absolute value of a m plus 1 divided by a m. So let's find our terms. Starting with the absolute value. Let's write down a n + 1. That'll be 2 to the power of n + 1 + 1. So that would be M + 2 in the exponent, right, divided by. Alan of And plus 1. And now we're dividing by AN. In other words, we can multiply by the reciprocal of it. So we're multiplying by ln of n. Divided by t to the power of n + 1. And now since n approaches positive infinity, our terms are going to be positive, so we can drop the absolute value and use the properties of exponents to simplify, so that would be 2 to the power of n + 1 multiplied by 2. Divided by ln of n + 1. Multiplied by ln of n divided by 2 to the power of n + 1, which allows us to cancel out 2 to the power of n + 1. And now we can rewrite our limit. L equals, let's factor out the constant, that's 2. Limit as n approaches infinity of the ratio between l n of n and ln of. M + 1 We can use laptop's rule to find this limit, right, because As an approaches infinity, the numerator and the denominator, they both approach infinity. So let's differentiate. We get to limit as n approaches infinity of 1 divided by n. And in the denominator we get 1 divided by n plus 1. Those are the required derivatives, right? So now let's Calculate the result that's to limit as n approaches infinity of n + 1. Divided by um. We can now divide, right, which is going to give us 1 + 1 divided by n, and the fractional term is going to approach 0 because we have a constant divided by infinity. So finally we get 2 multiplied by 1, which gives us a limit of 2, and our 2 is greater than 1. So according to the ratio test, if our limit is greater than 1, the series diverges. The correct answer is option B. Thank you for watching.

Using the Ratio TestIn Exercises 1–8, use the Ratio Test to determine whether each series converges absolutely or diverges.∑(from n=2 to ∞) [(3ⁿ⁺²) / ln(n)]

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Using the Ratio Test
In Exercises 1–8, use the Ratio Test to determine whether each series converges absolutely or diverges.
∑(from n=2 to ∞) [(3ⁿ⁺²) / ln(n)]