Explain why the integral test does not apply to the series.∑n=0∞nn2−1\sum_{n=0}^{\infty}\frac{n}{n^2-1}

Not continuous at x = 1 x=1 x = 1

Explain why the integral test does not apply to the series. ∑ n = 0 ∞ n n 2 − 1 \sum_{n=0}^{\infty}\frac{n}{n^2-1}

we are asked to explain why the integral test does not apply to the following series. And we have this series from n equals zero to infinity of n over n squared minus one. Now the first thing we should do when trying to solve this problem is figure out what our function is going to be, and that's gonna be based off this a sub n right here. So I'm gonna set that equal to f of x. Now my f of x here, well, that's just going to be taking this n here and replacing it with x. So we'll have x over x squared minus one. Now there are three criteria that need to apply in order to use the integral test. First off, the function that we have needs to be positive. Next, our function needs to be continuous, and then we need to have a case where this is going to be decreasing. So these are the three criteria that need to apply for the integral test, and this is going to be for x is greater than or equal to whatever value we start at right here, which in this case is zero. So first off, let's see if this function is indeed going to be positive. Well, I can see that we have this x squared minus one on bottom, and then we have this x on top here. Now what I would do is I'd try with starting with f of zero, and f of zero would give us zero over zero squared minus one. But this whole thing would just come out to zero. Now what about f of one? Well, f of one here, that's actually that's going to create a case where we have one squared minus one. So we'd end up with a one on top, but then we'd end up with one squared minus one, which is all going to be one over zero, which is not something that's defined. So So what I'll go ahead and do is skip past this, and let's try now f of two. Well, for f of two, we're going to have two over two squared minus one. This is the same thing as two over four minus one, which I can see will give me two thirds. If I go ahead and keep going here, notice I'm never going to get a negative value that comes out. So because of this, we can say that our function is positive for x is greater than or equal to zero. Now what I wanna do is move to this next piece right here because this is where things get interesting. In order for our function to be continuous, we can't have any place where this function breaks. But notice we already found a scenario right here where the function breaks. We get one over zero. This is undefined. And so because of this, we cannot say that our function is continuous for all x is greater than or equal to zero. So because of this, we cannot apply the integral test, this criteria is broken. So we would say this is not continuous at x is equal to one. And this right here would be our solution to the problem. So that is why we cannot use the integral test in this situation.

Explain why the integral test does not apply to the series. ∑ n = 0 ∞ n n 2 − 1 \sum_{n=0}^{\infty}\frac{n}{n^2-1}

Not continuous at x = 1 x=1 x = 1

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.
$\sum_{n = 0}^{\infty} \frac{n}{n^{2} - 1}$

The series is increasing

Not continuous at $x=1$

The series is decreasing

The series is not positive

Verified step by step guidance

Step 1: Recall the conditions for the integral test. The integral test applies to a series ∑aₙ if the corresponding function f(x) is positive, continuous, and decreasing for x ≥ 1.

Step 2: Analyze the given series ∑n=0∞n/(n²−1). The term aₙ = n/(n²−1) corresponds to the function f(x) = x/(x²−1).

Step 3: Check the continuity of f(x). The function f(x) = x/(x²−1) is undefined at x = 1 because the denominator becomes zero (x²−1 = 0). This means f(x) is not continuous at x = 1, violating one of the conditions for the integral test.

Step 4: Check if the series is positive. For x > 1, the numerator x is positive, but the denominator x²−1 is also positive. However, for x < 1, the denominator x²−1 becomes negative, making f(x) negative. Thus, the series is not positive for all x ≥ 1, violating another condition for the integral test.

Step 5: Check if the series is decreasing. While the series may appear to decrease for large values of n, the integral test requires the function to be strictly decreasing for all x ≥ 1. Since the series fails the other two conditions (continuity and positivity), the integral test cannot be applied regardless of whether the series is decreasing.

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