Problem 9 - Thomas Calculus

Question

Use the Integral Test to determine if the series in Exercises \(1-10\) converge or diverge. Be sure to check that the conditions of the Integral Test are satisfied. $$ \sum_{n=1}^{\infty} \frac{n^{2}}{e^{n / 3}} $$

Step-by-Step Solution

Answer

The series converges.

Step 1: Define the Function

Identify the function corresponding to the terms of the series. Here, the series is given by \( \sum_{n=1}^{\infty} \frac{n^{2}}{e^{n / 3}} \), so we define the function \( f(x) = \frac{x^{2}}{e^{x/3}} \).

Step 2: Verify the Conditions

Confirm that the conditions for the Integral Test are met. The function \( f(x) \) should be continuous, positive, and decreasing for \( x \geq 1 \). Since the exponential function grows faster than the polynomial, \( f(x) \) is positive and decreasing for \( x \geq 1 \).

Step 3: Set Up the Integral

To apply the Integral Test, we calculate the improper integral \( \int_1^{\infty} \frac{x^{2}}{e^{x/3}} \, dx \).

Step 4: Substitution Method

Use substitution to evaluate the integral. Let \( u = x/3 \), then \( du = 1/3 \, dx \), or \( dx = 3 \, du \). Change the limits of integration: when \( x = 1 \), \( u = 1/3 \); when \( x \to \infty \), \( u \to \infty \).

Step 5: Simplify the Integral

Substitute \( u \) into the integral:\[\int_{1/3}^{\infty} 3(3u)^2 e^{-u} \, du \]This simplifies to \( 27 \int_{1/3}^{\infty} u^{2} e^{-u} \, du \).

Step 6: Evaluate the Simplified Integral

The integral \( \int_{1/3}^{\infty} u^{2} e^{-u} \, du \) converges. This type of integral is a form of the gamma function, which converges for powers of \( u \) greater than \(-1\). Hence, the original integral converges.

Step 7: Conclude Based on the Integral Test

Since the integral \( \int_{1}^{\infty} \frac{x^{2}}{e^{x/3}} \, dx \) converges, by the Integral Test, the series \( \sum_{n=1}^{\infty} \frac{n^{2}}{e^{n/3}} \) also converges.

Key Concepts

Convergence of Series

When tackling series in mathematics, we are often interested in whether the series converges or diverges. **Convergence** means the series has a finite limit, while **divergence** implies it grows indefinitely. The **Integral Test** is a powerful tool for determining convergence or divergence of an infinite series.

To determine convergence using the Integral Test, you must first check if the function corresponding to the terms of the series is continuous, positive, and decreasing for all relevant values.

• The function must be continuous on the interval you are considering; any breaks or jumps disqualify it from the test.
• The function must be positive, ensuring the terms contribute meaningfully to the growth or convergence.
• It should be decreasing, which helps confirm that the series's terms diminish as they progress.

By ensuring these conditions, you can then use the test to relate the sum of the series to a corresponding improper integral.

Improper Integrals

Integral calculus allows us to compute area under curves, but sometimes this calculation stretches over an infinite range or involves unbounded behavior. These are called **improper integrals**. They are slightly different from regular integrals since they can extend to infinity or encompass points where the function returns infinity.

For example, consider the integral \( \int_1^{\infty} \frac{x^{2}}{e^{x/3}} \, dx \). This is improper because of the infinite upper limit. When evaluating such integrals, we often use limits to define them properly:

• Transform the integral into a proper form — \( \lim_{b \to \infty} \int_1^b \frac{x^{2}}{e^{x/3}} \, dx \).
• Determine if this limit exists (finite number) or diverges (goes to infinity).

If the limit is finite, we have a convergent integral, suggesting the behavior of the series is also convergent.

Gamma Function

In mathematical analysis, the **Gamma Function** is a versatile function that extends the concept of factorial to non-integer values. It appears frequently in evaluations of improper integrals, especially when the integral involves exponential decay combined with polynomial terms. In our series, for instance, the evaluated integral \( \int_{1/3}^{\infty} u^{2} e^{-u} \, du \) is a form related to the Gamma Function.

The Gamma Function, denoted \( \Gamma(n) \), for integer \( n \) coincides with the factorial, i.e., \( \Gamma(n) = (n-1)! \). It can be represented for complex argument \( z \) as: \[ \Gamma(z) = \int_0^{\infty} t^{z-1} e^{-t} \, dt \]
This connection makes it a key player in determining convergence as many convergent integrals can be expressed in terms of the Gamma Function when they fit the form \( \int_{a}^{b} x^{n} e^{-x} \, dx \).

• If the power of \( x \) in your integral is positive (greater than \(-1\)), the Gamma Function ensures convergence.
• This function simplifies manual integration by shifting the work into tabulated or computational resources.

Understanding and applying the Gamma Function can bridge gaps when evaluating complex situations in calculus.