Question

Use the Integral Test to determine the convergence of the given series. $$\sum_{n=1}^{\infty} \frac{n}{2^{n}}$$

Short answer

The series \( \sum_{n=1}^{\infty} \frac{n}{2^{n}} \) converges.

Step-by-step solution

Identify the Function

To apply the Integral Test, we need to identify a function that represents the terms of the series. For the series \( \sum_{n=1}^{\infty} \frac{n}{2^{n}} \), we can consider the function \( f(x) = \frac{x}{2^{x}} \). This choice is natural given the structure of the terms in the series.

Check Conditions of the Integral Test

The Integral Test applies if the function \( f(x) = \frac{x}{2^x} \) is positive, continuous, and decreasing for \( x \geq 1 \). Since \( f(x) \) involves exponential growth in the denominator, which outpaces polynomial growth in the numerator, \( f(x) \) is indeed positive and continuous. To show it's decreasing, we check its derivative: \( f'(x) = \frac{2^x - x \cdot 2^x \ln(2)}{(2^x)^2} = \frac{2^x (1 - x \ln(2))}{2^{2x}} \), which is negative for \( x \geq 1 \).

Evaluate the Integral

Integrate \( f(x) = \frac{x}{2^x} \) from 1 to \( \infty \). Consider the integral \( \int_{1}^{\infty} \frac{x}{2^x} \, dx \), which can be found using integration by parts. Let \( u = x \) and \( dv = \frac{1}{2^x} \, dx = (\frac{1}{\ln(2)})(-e^{-x \ln(2)}) \, dx \). Solve using integration by parts formula: \( \int u \, dv = uv - \int v \, du \).

Apply Integration by Parts

For integration by parts, \( u = x \), \( du = dx \), \( dv = \frac{2^{-x}}{\ln(2)} \, dx \), and \( v = -\frac{1}{2^{x} \ln(2)} \). Thus, the integral becomes: \[ x \left(-\frac{1}{2^x \ln(2)} \right) \Big|_{1}^{\infty} - \int_{1}^{\infty} \left(-\frac{1}{2^x \ln(2)} \right) dx \]. We separate and solve each part to find convergence.

Conclusion: Evaluate Limits

After integration, evaluate the limit: \( \lim_{b \to \infty} \left[-\frac{x}{2^x\ln(2)}\right]_1^b + \frac{1}{\ln(2)}\left[\frac{1}{2^x \ln(2)}\right]_1^b \). For the first term, as \( x \to \infty \), the expression goes to zero due to the exponential denominator. For the second term: \( \frac{1}{2^x} \to 0 \) as \( x \to \infty \). This confirms that the integral converges.

Key concepts

Understanding Series Convergence

When tackling the convergence of a series, it's essential to determine whether the sum of its terms approaches a finite number or not. The series in question is \( \sum_{n=1}^{\infty} \frac{n}{2^{n}} \). This specific series can be analyzed using the Integral Test, a powerful tool for assessing series convergence.

The Integral Test involves checking if a corresponding function \( f(x) = \frac{x}{2^{x}} \) meets three key criteria:

• The function must be positive for \( x \geq 1 \).
• It should be continuous over its domain.
• The function needs to be decreasing for \( x \geq 1 \).

If these conditions hold, the convergence of the series can be determined by evaluating the improper integral \( \int_{1}^{\infty} f(x) \, dx \). If the integral converges, so does the series. In our case, thanks to the rapid growth of the exponential term in the denominator, the function \( f(x) \) is positive, continuous, and decreasing for \( x \geq 1 \). Thus, the series' convergence can be found by applying the Integral Test.

Exploring Integration by Parts

Integration by parts is a technique used to solve integrals involving products of functions. When applying the Integral Test to the series \( \sum_{n=1}^{\infty} \frac{n}{2^{n}} \), we encounter the integral \( \int_{1}^{\infty} \frac{x}{2^x} \, dx \). Integration by parts becomes the method of choice here due to the form of the integrand.

The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \). For our integral, let:

• \( u = x \), which means \( du = dx \).
• \( dv = \frac{1}{2^x} \, dx = \frac{e^{-x \ln(2)}}{\ln(2)} \, dx \), leading to \( v = -\frac{1}{2^x \ln(2)} \).

Substituting these into the formula allows us to evaluate the integral's convergence. We compute two distinct parts: the product \( uv \) evaluated at the relevant bounds, and the remaining integral \( \int v \, du \). This approach simplifies the process of handling complex integrals, especially when exponential and polynomial terms intertwine.

The Role of Exponential Functions

Exponential functions play a crucial role in maintaining the structure and behavior of many series, such as \( \sum_{n=1}^{\infty} \frac{n}{2^{n}} \). In these functions, a variable appears as an exponent, often leading to rapid growth or decay.

For the series mentioned above, the exponential term \( 2^n \) in the denominator grows significantly faster than the linear term \( n \) in the numerator. This disparity ensures the function \( f(x) = \frac{x}{2^x} \) decreases as \( x \) rises, which is a condition of the Integral Test. Essentially, the rapid increase of the denominator diminishes the size of the terms in the series, pivotal in deducing convergence through integration.

• Exponential growth often means fast-growing tails in a sequence or a function.
• The balance between growth rates determines the overall behavior of the series.
• Utilizing exponential functions in analysis requires understanding their asymptotic behavior.

Thus, in mathematical terms, exponential functions are indispensable in predicting and explaining how series behave based on their convergence characteristics.