# Question Video: Determine Whether a Series Is Convergent or Divergent Using the Integral Test Mathematics • Higher Education

Using the integral test, determine whether the series 1/(2ln(2)) + 1/(3ln(3)) + 1/(4ln(4)) + βββ is convergent or divergent.

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### Video Transcript

Using the integral test, determine whether the series one over two multiplied by the natural log of two add one over three multiplied by the natural log of three add one over four multiplied by the natural log of four and so on is convergent or divergent.

Letβs begin by reminding ourselves of what the integral test is. Suppose that π of π₯ is continuous, positive, and decreasing for all values of π₯ greater than or equal to some value π and π of π equals π π. Then if the integral between π and β of π of π₯ with respect to π₯ is convergent, then so is the series the sum from π equals π to β of π π. If the integral between π and β of π of π₯ with respect to π₯ is divergent, then so is the series the sum from π equals π to β of π π.

Note that both of these integrals are improper integrals. Thatβs because one of the limits of integration is β. We deal with improper integrals by replacing the infinite limit with a variable and then evaluating the integral as that variable approaches β. If we then find that when we evaluate the integral we get β, then the integral is divergent. Otherwise, if we get a finite number, then the integral is convergent.

Letβs start by finding a way to express this series in Ξ£ notation. We can see that it has the form the sum of one over π multiplied by the natural log of π. But weβve got to spot that this runs from π equals two to β. So our π π is one over π multiplied by the natural log of π. So the function π of π₯ that weβre going to be integrating is one over π₯ multiplied by the natural log of π₯. And our limits of integration are two and β. In order to use the integral test, we must check that weβve met the conditions that π of π₯ is continuous, positive, and decreasing on the interval.

So letβs start with checking if this function π of π₯ is continuous. We need to make sure that this is continuous from π₯ is greater than or equal to two. And because the quotient of two continuous functions is continuous and π₯ multiplied by the natural log of π₯ is defined for all π₯ greater than or equal to two and is not equal to zero on our interval, then one over π₯ multiplied by the natural log of π₯ is continuous. We also have that all values of one over π₯ multiplied by the natural log of π₯ are positive for π₯ greater than or equal to two. And finally, as the natural log of π₯ is an increasing function, so is π₯ multiplied by the natural log of π₯. So one over π₯ multiplied by the natural log of π₯ is a decreasing function. So weβve satisfied all of the conditions required for the integral test.

So remember weβre trying to determine the convergence of this integral. And as this is an improper integral, weβre going to rewrite this as a limit. This is the limit as π‘ approaches β of the integral between two and π‘ of one over π₯ multiplied by the natural log of π₯ with respect to π₯.

Letβs evaluate this integral using integration by substitution. If we start with the indefinite integral one over π₯ multiplied by the natural log of π₯ with respect to π₯, then using the substitution π’ equals the natural log of π₯ so that dπ’ by dπ₯ equals one over π₯. And remember that dπ’ by dπ₯ is not a fraction. But we do manipulate it a little bit like a fraction when weβre doing integration by substitution. So this rearranges to give us that dπ’ equals one over π₯ dπ₯. Then we can replace one over the natural log of π₯ with one over π’. And we can replace the one over π₯ dπ₯ with dπ’.

At this point, we recall the general result that tells us that the integral of one over π’ with respect to π’ is the natural log of the absolute value of π’. And because here weβre doing indefinite integration, we add a constant of integration π. But because weβre actually looking for the integral between two and π‘ of one over π₯ multiplied by the natural log of π₯ with respect to π₯, we need to substitute back in π’ equals the natural log of π₯. This gives us the limit as π‘ approaches β of the natural log of the absolute value of the natural log of π₯.

And remember we need to evaluate this between two and π‘. This gives us the limit as π‘ approaches β of the natural log of the absolute value of the natural log of π‘ minus the natural log of the absolute value of the natural log of two. But the natural log of the absolute value of the natural log of π‘ approaches β. So evaluating this whole expression with the limit as π‘ approaches β just gives us β. This result means that the improper integral between two and β of one over π₯ multiplied by the natural log of π₯ with respect to π₯ is divergent. And the integral test tells us that if we find this integral to be divergent, then this series is also divergent. So by the integral test, this series is divergent.