Explain why the integral test does not apply to the series.∑n=1∞(−1)nlnn\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}}{\ln n}

f ( x ) f\left(x\right) is not always positive.

Explain why the integral test does not apply to the series. ∑ n = 1 ∞ ( − 1 ) n ln n \sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}}{\ln n}

this problem, we are asked to explain why the integral test does not apply to the following series. Now the series that we have here is the sum from n equals one to infinity of negative one to the n power over the natural log of n. So how can I go about solving this problem? Well, this is kind of an interesting case where we have to figure out why the integral test does not apply. And the way that I can do this is by thinking about the criteria that need to be met for the integral test. Now the way that we can find these criteria is by first off figuring out what function it is that we're dealing with. And the function is going to be based on this piece, a sub n. So I'm gonna take my a sub n instead of equal to my function f of x, meaning I need to take all of my n's here and replace them with x. So that's gonna give me the function f of x is equal to negative one to the x power over the natural log of x. Now when it comes to the integral test, there are three criteria that need to be met. First off, your function has to be positive. Secondly, your function has to be continuous, and then you have to have that your function is decreasing for all values that you're looking at on your interval. Now the interval I can see is going to start from n equals one here, so and it goes to infinity, so we're going to say that this is for x that is greater than or equal to one. Now first off, we should check if this is going to be positive. Now what I can see based on the function we have is we have this negative one to the x power right here. If we have negative one to the x, this is actually going to make our function sometimes negative and sometimes positive. It depends on what this x is. Because if we had negative one to the one power, that would give us something negative. If we had negative one squared, that gives us negative one times negative one, so the negatives cancel giving us positive one. Negative one cubed, well, that would give us back to negative one, and then negative one to the fourth, well, that's gonna get us back to positive one. So notice that we're just gonna keep going back and forth between negative, positive, negative, positive. So we cannot say that this function is going to be positive for all cases where x is greater than or equal to one. So because of that, this criteria is not met, and that is why the integral test does not apply. So to write a solution for this, what we would say is that our function f of x is not always positive, because we can see here that this is not going to be the case because it alternates in sign. So in this case, this would be our solution, and that is why we cannot use the integral test.

Explain why the integral test does not apply to the series. ∑ n = 1 ∞ ( − 1 ) n ln ⁡ n \sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}}{\ln n}

f ( x ) f\left(x\right) is not always positive.

f ( x ) f\left(x\right) f ( x ) is always positive.

f ( x ) f\left(x\right) f ( x ) is not continuous.

f ( x ) f\left(x\right) f ( x ) is decreasing.

14. Sequences & Series

Convergence Tests

Calculus

14. Sequences & Series

Convergence Tests

Struggling with Calculus?

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Multiple Choice

Explain why the integral test does not apply to the series.
$\sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{\ln n}$

$f\left(x\right)$ is always positive.

$f\left(x\right)$ is not continuous.

$f\left(x\right)$ is decreasing.

$f (x)$ is not always positive.

Verified step by step guidance

The integral test is a method used to determine the convergence of an infinite series by comparing it to the improper integral of a related function. For the integral test to apply, the function f(x) corresponding to the terms of the series must satisfy three conditions: it must be positive, continuous, and decreasing for all x greater than or equal to some value.

In the given series, the general term is (-1)^n / ln(n). To apply the integral test, we consider the function f(x) = (-1)^x / ln(x).

The first condition for the integral test is that f(x) must be positive. However, the term (-1)^x introduces oscillation between positive and negative values depending on whether x is an integer and whether x is odd or even. This means f(x) is not always positive.

Since f(x) is not always positive, the integral test cannot be applied to this series. The positivity condition is violated, which is a fundamental requirement for the integral test.

In conclusion, the integral test does not apply to the series because the function f(x) = (-1)^x / ln(x) is not always positive, as required by the test.

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