10.4 Integral Test for Convergence

10.4 Integral Test for Convergence

Welcome to AP Calc 10.4! In this guide, you’ll learn how to apply another test to determine whether series are convergent or divergent. This test will rely heavily on your understanding of indefinite integrals from Unit 6.

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∫ Integral Test Theorem

The integral test states that, for a positive, decreasing function $f(x)$ over the interval $[k,\infty)$ with a corresponding sequence $a_n=f(x)$, then…

(1) if $\int_k^{\infty}\ f(x)\ \text{d}x$ converges, then $\sum_{n=k}^{\infty}a_n$ also converges,

(2) if $\int_k^{\infty}\ f(x)\ \text{d}x$ diverges, then $\sum_{n=k}^{\infty}a_n$ also diverges,

and…

$$a_k+\int_{k+1}^{\infty}\ f(x)\ \text{d}x\leq \sum_{n=k}^{\infty}a_n\leq \int_k^{\infty}\ f(x)\ \text{d}x$$

For now, we can ignore the comparison portion and just focus on the first points! Let’s break down what everything means.

🧱 Breaking Down the Integral Test Theorem

First, we need a positive, decreasing function. This means it must be above zero over the entire interval, and decreasing, or getting smaller over time. Consider the function $f(x)=\frac{1}{x\ \text{ln}\ (x)}$ over the interval $[2,\infty)$:

Image of the function $1/(xlnx)$ over the interval $2,15]$. Image courtesy of [Desmos

This function satisfies our initial condition. Now, we just need a series where $a_n=f(x)$. This is pretty easy to set up!

$$\sum_{n=2}^{\infty}\frac{1}{n\ \text{ln}(n)}$$

Now, we can apply our test!

✏️ Applying the Integral Theorem

To apply our test, we first have to take the integral of our function:

$$\int_2^{\infty} \frac{1}{x\ \text{ln}\ (x)}\ \text{d}x$$

To solve this integral, we need to apply $u$-substitution, where $u=\text{ln}\ x$ and $\text{d}u=\frac{1}{x}\ \text{d}x$:

$$\int \frac{1}{u}\ \text{d}u=\text{ln} (u)$$

Undoing $u$-substitution, we find that

$$\int_2^{\infty} \frac{1}{x\ \text{ln}\ (x)}\ \text{d}x=\text{ln}(|\text{ln}(x)|)+C$$

Finally, we evaluate over our bounds:

$$\text{ln}(|\text{ln}(\infty)|)-\text{ln}(|\text{ln}(2)|)=\infty$$

Because this integral evaluates as $\infty$, we say it is divergent. Based on our theorem, this means that the series $\sum_{n=2}^{\infty}\frac{1}{n\ \text{ln}(n)}$ is also divergent.

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📝 Integral Test Theorem Practice

Now it’s your turn to apply what you’ve learned!

❓Integral Test Theorem Problems

1. State the integral test and its conditions.

2. Use the integral test to determine whether the series diverges.

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

3. Use the integral test to determine whether the series diverges.

   $$\sum_{n=0}^{\infty}\frac{n^3}{n^4+1}$$

💡 Integral Test Theorem Solutions

1. Review the beginning of the guide for the full theorem! You need a function that is decreasing and positive over an interval and a series using that function.

2. The integral $\int_{0}^{\infty}\frac{1}{1+x^2} \ \text{d}x$ is a standard integral, giving $[\text{arctan}(x)]_0^{\infty}$ which gives $\pi/2-0=\pi/2$, which means that the series is convergent.

3. The integral $\int_{0}^{\infty}\frac{x^3}{x^4+1}\ \text{d}x$ is evaluated using $u$-substitution, where $u=x^4+1$ and $\text{d}u=4x^3\ \text{d}x$:

   $$\frac{1}{4}\int\frac{1}{u}\ \text{d}u=\frac{\text{ln}(u)}{4}$$

   Undoing $u$-substitution, we get:

   $$\Bigg[\frac{\text{ln}(x^4+1)}{4}\Bigg]_0^\infty=\infty-0=\infty$$

   Because the integral is not finite, we say it diverges. Therefore, our series diverges.

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   Excellent job! Be sure to practice this type of problem and make sure you know the conditions for using this test. Once you’ve got it down, the integral test will be a breeze! 💯