Use the Integral Test to determine whether the infinite...

Use the Integral Test to determine whether the infinite series is convergent. [ sum_{n=1}^{infty} frac{2}{3^{ln n}} ] Fill in the corresponding integrand and the value of the improper integral. Enter inf for ( infty,-inf ) for ( -infty ), and DNE if the limit does not exist. Compare with ( int_{1}^{infty} frac{2}{3^{ln x}} quad d x= ) By the Integral Test, the infinite series ( sum_{n=1}^{infty} frac{2}{3^{ln n}} ) A. converges B. diverges

Transcript

00:01 Okay, we're looking how to compute the indefinite integral here, which is the limit as a goes to infinity of this integral from 1 to a of 2 over 3 to the natural log x dx.

00:18 Well, it's pretty clear here that the method we should use is a u substitution because there's one particular thing that's making this integral difficult and that's that natural log x.

00:27 And so let's let you equal l and x, which means that du equals 1 over x, which 1 over x, d x, which is going to be a pesky thing to get rid of.

00:38 So how we're going to rewrite this is we're going to pull the two on the outside.

00:41 This will be the limit as a goes to infinity.

00:44 When x equals 1, u equals 0, when x equals a, you equals lna.

00:57 And then we have originally, this would be 2 times 1 over 1.

01:04 3 natural log x which would be just 1 over 3 to the u but d u is equal to 1 over x d x and so we need to figure out a way to get rid of this 1 over x and so we're going to end up having this be e to the u because that would be an x on top d u so that e to the u over 3 to the u d u this is equal to x over 3 to the u d u this is equal to x over 3 to the log x and d u is one over x d x and the x is cancel and where we have verified our u substitution so we can rewrite this as the limit as a goes to infinity of two times the integral from zero to natural log a of e over three to the power u to you and we know explicitly we can evaluate the integral of any a to the power u um let me just check my notes that that's just a to the x over l and a.

02:21 So this is the limit as a goes to infinity.

02:26 Let's pull the two outside again.

02:30 Of e over 3 to the u divided by natural log of u.

02:43 And this is all evaluated from u equals lna to u equals 0.

02:51 So the u equals zero case is just a constant.

02:54 So we're really concerned with what is the limit as a goes to infinity of e over 3 to the natural log of a.

03:27 Well, let's do some checking real quick...

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