The Integral Test is a really helpful method when studying series. It’s especially useful for figuring out whether infinite series, which go on forever, are converging (settling down to a number) or diverging (growing without limit). By connecting series and integrals, we can analyze series that seem hard to understand at first.

How Series and Improper Integrals are Related

First, let’s look at how series and improper integrals connect. An infinite series can be written as:

$$\sum_{n=1}^{\infty} a_n$$

We can compare this to an improper integral of a continuous, positive, and decreasing function. The terms in the series can be seen as the approximate areas under the curve of the function, which we call:

$$f(x) = a_n \quad \text{for } n = 1, 2, 3, \ldots$$

The integral we look at is:

$$\int_1^{\infty} f(x) \, dx.$$

We call this integral "improper" because it goes on to infinity. By seeing if this integral converges or diverges, we can understand the behavior of the series. If the integral converges, then the series does too. If the integral diverges, then the series diverges as well.

Conditions for Using the Integral Test

For the Integral Test to work, a few conditions need to be met:

1. Positive: For all $x \geq 1$, the function must be positive, which means $f(x) > 0$.

2. Continuous: The function should be continuous for the range from $[1, \infty)$.

3. Decreasing: The function must be decreasing. This means, if $x_1$ is greater than $x_2$ (both at least 1), then $f(x_1)$ must be less than $f(x_2)$.

When these conditions are met, we can use the Integral Test to analyze the series $\sum_{n=1}^{\infty} a_n$.

Using the Integral Test to Determine Series Convergence

Now, let’s see an example of how to use the Integral Test to figure out if an infinite series converges or diverges.

Take the series:

$$\sum_{n=1}^{\infty} \frac{1}{n^p}$$

where $p$ is a positive number. To check if it converges, we create the function:

$$f(x) = \frac{1}{x^p}.$$

Next, we check the three conditions:

1. Positive: For $x \geq 1$, $f(x) = \frac{1}{x^p} > 0$.

2. Continuous: The function $\frac{1}{x^p}$ is continuous for $x \geq 1$.

3. Decreasing: The function is decreasing because as $x$ gets bigger, $f(x)$ gets smaller.

Since $f(x)$ meets all the conditions, we can evaluate the improper integral:

$$\int_1^{\infty} \frac{1}{x^p} \, dx.$$

To solve this integral, we write:

$$\int_1^{\infty} \frac{1}{x^p} \, dx = \lim_{t \to \infty} \int_1^{t} \frac{1}{x^p} \, dx.$$

This integral gives us:

$$\int_1^{t} \frac{1}{x^p} \, dx = \left[ \frac{x^{1-p}}{1-p} \right]_1^{t} = \frac{t^{1-p}}{1-p} - \frac{1}{1-p} \quad (p \neq 1).$$

Now, we look at the limit as $t \to \infty$:

• If $p > 1$: The term $t^{1-p} \to 0$. So, the integral converges to

$$\frac{1}{p-1}.$$

• If $p = 1$: The integral simplifies to

$$\int_1^{t} \frac{1}{x} \, dx = \ln(t) - \ln(1) = \ln(t) \to \infty \quad \text{as } t \to \infty.$$

• If $0 < p < 1$: The term $t^{1-p} \to \infty$, so the integral diverges.

From the Integral Test, we can summarize:

• The series $\sum_{n=1}^{\infty} \frac{1}{n^p}$ converges if $p > 1$.
• The series diverges if $p \leq 1$.

Conclusion

The Integral Test is a smart and useful way to work with infinite series. It helps us see if a series converges or diverges by comparing it to an improper integral. Understanding this connection helps students and learners tackle tough series with confidence. The strong link between series and calculus not only improves our calculation skills but also deepens our understanding of convergence in math.