Integral Test Determine Convergence And Divergence Of Series

by JurnalWarga.com 61 views
Iklan Headers

Hey guys! Let's tackle the integral test today, a super handy tool in our calculus toolbox for figuring out if an infinite series converges or diverges. We'll break it down step by step and make sure you're feeling confident about using it.

The integral test is a powerful method used in calculus to determine the convergence or divergence of an infinite series by comparing it to an improper integral. This test is particularly useful when dealing with series whose terms can be expressed as a function of the index variable, allowing us to leverage the techniques of integral calculus.

The integral test bridges the gap between discrete sums and continuous integrals. The idea behind the integral test is quite intuitive. If we have a series $\sum_{n=1}^{\infty} a_n$, we can think of the terms $a_n$ as the values of a function $f(x)$ at integer values of $x$. If this function is positive, continuous, and decreasing on the interval $[1, \infty)$, then we can compare the sum of the series to the integral of the function over the same interval. Geometrically, we're comparing the sum of the areas of rectangles (representing the series) to the area under the curve (representing the integral).

If the improper integral $\int_1^{\infty} f(x) dx$ converges, meaning its value is a finite number, then the infinite series $\sum_{n=1}^{\infty} a_n$ also converges. Conversely, if the integral diverges, meaning it has an infinite value, then the series also diverges. This direct relationship allows us to use the convergence or divergence of the integral as a proxy for the convergence or divergence of the series.

To successfully apply the integral test, there are three key conditions that must be met by the function $f(x)$ corresponding to the terms of the series $a_n$:

  1. Positive: The function $f(x)$ must be positive for all $x$ greater than or equal to 1. This ensures that the terms of the series are positive, which is a prerequisite for the test.
  2. Continuous: The function $f(x)$ must be continuous on the interval $[1, \infty)$. This is necessary because we need to be able to evaluate the integral of the function over this interval.
  3. Decreasing: The function $f(x)$ must be decreasing for $x$ greater than or equal to 1. This condition is crucial because it ensures that the terms of the series are getting smaller as $n$ increases, which is a key factor in determining convergence. This means that as x gets larger, the value of f(x) should get smaller or, at the very least, not increase. Mathematically, this condition is satisfied if the derivative of $f(x)$, denoted as $f'(x)$, is negative for $x \geq 1$.

If these three conditions are satisfied, the integral test provides a powerful tool for determining the convergence or divergence of the series. It's important to verify each condition carefully before applying the test to ensure the validity of the conclusion.

Let’s dive into a specific example to see how the integral test works in practice. We'll investigate the series:

∑n=1∞47n−1\sum_{n=1}^{\infty} \frac{4}{7n-1}

Our mission is to determine whether this series converges or diverges using the integral test. To do this, we'll follow a structured approach, carefully checking each condition and then evaluating the corresponding integral.

Step 1: Define the Function

First, we need to define a continuous function $f(x)$ that corresponds to the terms of the series. We replace $n$ with $x$ in the general term of the series to obtain:

f(x)=47x−1f(x) = \frac{4}{7x-1}

This function will serve as the basis for our integral test. Now that we have our function, we need to verify that it meets the three essential conditions for the integral test: positivity, continuity, and decreasing behavior.

Step 2: Verify the Conditions for the Integral Test

Condition 1: Positivity

We need to ensure that $f(x) = \frac{4}{7x-1}$ is positive for all $x \geq 1$. For $x \geq 1$, the denominator $7x - 1$ is always positive since $7(1) - 1 = 6 > 0$, and it increases as $x$ increases. The numerator, 4, is also positive. Therefore, the fraction $\frac{4}{7x-1}$ is positive for all $x \geq 1$. So, the first condition is met.

Condition 2: Continuity

Next, we check for continuity. The function $f(x) = \frac{4}{7x-1}$ is a rational function, and rational functions are continuous everywhere except where the denominator is zero. The denominator $7x - 1$ is zero when $x = \frac{1}{7}$. Since we are considering the interval $[1, \infty)$, and $\frac{1}{7}$ is not in this interval, the function is continuous on $[1, \infty)$. Thus, the second condition is satisfied.

Condition 3: Decreasing

To verify that $f(x)$ is decreasing, we can examine its derivative, $f'(x)$. Let's find the derivative of $f(x)$$

f(x)=4(7x−1)−1f(x) = 4(7x - 1)^{-1}

f′(x)=−4(7x−1)−2imes7=−28(7x−1)2f'(x) = -4(7x - 1)^{-2} imes 7 = \frac{-28}{(7x - 1)^2}

Since the numerator is negative and the denominator is always positive for $x \geq 1$, the derivative $f'(x)$ is negative for all $x \geq 1$. This means that the function $f(x)$ is decreasing on the interval $[1, \infty)$. So, the third condition is also met.

Since all three conditions (positive, continuous, and decreasing) are satisfied, we can confidently apply the integral test to determine the convergence or divergence of the series.

Step 3: Evaluate the Improper Integral

Now that we've confirmed the conditions, let's evaluate the improper integral:

∫1∞47x−1dx\int_1^{\infty} \frac{4}{7x-1} dx

To evaluate this improper integral, we'll use a limit:

∫1∞47x−1dx=lim⁡t→∞∫1t47x−1dx\int_1^{\infty} \frac{4}{7x-1} dx = \lim_{t \to \infty} \int_1^{t} \frac{4}{7x-1} dx

Let's find the antiderivative of $\frac{4}{7x-1}$. We can use a simple u-substitution. Let $u = 7x - 1$, so $du = 7 dx$, and $dx = \frac{1}{7} du$. The integral becomes:

∫47x−1dx=∫4u17du=47∫1udu=47ln⁡∣u∣+C=47ln⁡∣7x−1∣+C\int \frac{4}{7x-1} dx = \int \frac{4}{u} \frac{1}{7} du = \frac{4}{7} \int \frac{1}{u} du = \frac{4}{7} \ln|u| + C = \frac{4}{7} \ln|7x-1| + C

Now we can evaluate the definite integral:

lim⁡t→∞∫1t47x−1dx=lim⁡t→∞[47ln⁡∣7x−1∣]1t\lim_{t \to \infty} \int_1^{t} \frac{4}{7x-1} dx = \lim_{t \to \infty} \left[ \frac{4}{7} \ln|7x-1| \right]_1^{t}

=lim⁡t→∞(47ln⁡∣7t−1∣−47ln⁡∣7(1)−1∣)= \lim_{t \to \infty} \left( \frac{4}{7} \ln|7t-1| - \frac{4}{7} \ln|7(1)-1| \right)

=lim⁡t→∞(47ln⁡(7t−1)−47ln⁡(6))= \lim_{t \to \infty} \left( \frac{4}{7} \ln(7t-1) - \frac{4}{7} \ln(6) \right)

As $t$ approaches infinity, $\ln(7t-1)$ also approaches infinity. Therefore:

lim⁡t→∞47ln⁡(7t−1)=∞\lim_{t \to \infty} \frac{4}{7} \ln(7t-1) = \infty

Thus, the improper integral diverges:

∫1∞47x−1dx=∞\int_1^{\infty} \frac{4}{7x-1} dx = \infty

Step 4: Conclusion

Since the improper integral $\int_1^{\infty} \frac{4}{7x-1} dx$ diverges, by the integral test, the series $\sum_{n=1}^{\infty} \frac{4}{7n-1}$ also diverges.

So there you have it! By carefully applying the integral test, we've determined that the given series diverges. Remember, the key is to check the conditions and then evaluate the integral. Keep practicing, and you'll master this technique in no time!

So, to wrap things up, the integral test is your friend when you need to figure out if a series converges or diverges. Just remember to check those three conditions – positive, continuous, and decreasing – and then get integrating! You've got this, and happy calculating!