Problem 17 - Essential Calculus

Question

Use the power series $$\frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^{n} x^{n}$$ to determine a power series, centered at 0 , for the function. Identify the interval of convergence. $$ f(x)=-\frac{1}{(x+1)^{2}}=\frac{d}{d x}\left[\frac{1}{x+1}\right] $$

Step-by-Step Solution

Answer

The required power series for the function \(f(x)\) is \(\sum_{n=1}^{\infty}(n)(-1)^{n}x^{n-1}\). The interval of convergence is \( -1

Step 1: Identify Given Power Series

The given power series is \( \frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^{n}x^{n} \). The function \(f(x)\) can be expressed as the derivative of the given power series, which means that we need to take the derivative of the right side of the equation.

Step 2: Deriving the Power Series

By taking the derivative of \(\sum_{n=0}^{\infty}(-1)^{n}x^{n}\), we get the following series: \(\sum_{n=1}^{\infty}(n)(-1)^{n}x^{n-1}\). Notice that the result looks almost like the original series, but multiplied by \( n \) and the power of \(x\) decreases by 1.

Step 3: Calculate Interval of Convergence

For the series \(\sum_{n=0}^{\infty}(-1)^{n}x^{n}\), the interval of convergence is \( -1

Step 4: Checking Endpoint Convergence

When \(x=-1\), the series becomes \(\sum_{n=1}^{\infty}(n)(-1)^{n-1}\), which does not converge. Similarly, when \(x=1\), the series becomes \(\sum_{n=1}^{\infty}(n)(-1)^{n}\), which also does not converge.

Key Concepts

Interval of Convergence

The interval of convergence is a fundamental concept when dealing with power series. It describes the range of values for which the series converges to a finite sum. For the power series \( \sum_{n=0}^{\infty}(-1)^{n} x^{n} \), the interval of convergence is \(-1 < x < 1\). That means within this range, the series will sum to a finite value.

When a power series is differentiated, the interval of convergence usually stays the same. However, one must be cautious about the endpoints of the interval. For the derived series \( \sum_{n=1}^{\infty}(n)(-1)^{n}x^{n-1} \), the same interval \(-1
The endpoints of the interval, \( x = -1 \) and \( x = 1 \), need special attention. Testing these can reveal if the series actually manages to converge at the boundary points. In our case, testing revealed non-convergence at both endpoints.

Here’s a simple checklist when identifying intervals of convergence:

• Evaluate the basic interval from the original series.
• Apply the same interval to any resulting derived series.
• Assess convergence at each endpoint.

Derivative of a Series

Taking the derivative of a power series follows a consistent pattern. Consider the power series \( \sum_{n=0}^{\infty}(-1)^{n} x^{n} \), derived from the function \( f(x) = \frac{1}{x+1} \). You can think of each term individually when differentiating.

The rule here is straightforward: differentiate each term in the series. The exponent of \( x \) decreases by one, and we multiply by the current exponent value \( n \). This transforms the series as follows: \( \sum_{n=1}^{\infty}(n)(-1)^{n}x^{n-1} \).

Here are some quick tips for differentiating power series:

• Remember to shift the index if needed (in this case, starting from \( n=1\) rather than \( n=0\)).
• Decrease the power of \( x \) and multiply by the original power (or index number \( n \)).
• Don't forget that differentiating a series can change the way terms are defined, but typically maintains the original convergence interval.

Convergence Tests

Convergence tests are mathematical tools that help verify if terms in a series converge to a sum. Applying these tests is crucial, especially at the endpoints of convergence intervals.

For a given power series, like \( \frac{1}{1+x} = \sum_{n=0}^{\infty}(-1)^{n}x^{n} \), we naturally use these tests to check if the behavior at \( x = -1 \) and \( x = 1 \) fits within our convergence range.

Common tests include:

• Ratio Test: This checks the absolute limit of successive terms. If the limit is less than one, the series converges absolutely. Not particularly applicable for endpoints here.
• Alternating Series Test: Useful for series that alternate in sign. It ensures that the terms are decreasing in absolute value for convergence.
• Comparison Test: Compare with a known convergent series to establish convergence by similarity.

In our example, checking convergence at the endpoints \( x=-1 \) and \( x=1 \) through simple observation showed divergence. This confirmed that our series remains non-convergent at these boundary conditions.

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Problem 17

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Problem 18

Other exercises in this chapter

Problem 17

Use the binomial series to find the Maclaurin series for the function. $$ f(x)=\frac{1}{\sqrt{4+x^{2}}} $$

View solution

Problem 17

In Exercises \(17-20\), approximate the sum of the series by using the first six terms. (See Example 4.) $$ \sum_{n=1}^{\infty} \frac{(-1)^{n+1} 3}{n^{2}} $$

View solution

Problem 18

Write the next two apparent terms of the sequence. Describe the pattern you used to find these terms. \(1,-\frac{1}{2}, \frac{1}{4},-\frac{1}{8}, \ldots\)

View solution

Problem 18

Explain why the Integral Test does not apply to the series. $$ \sum_{n=1}^{\infty}\left(\frac{\sin n}{n}\right)^{2} $$

View solution

Key Concepts

Interval of Convergence

The interval of convergence is a fundamental concept when dealing with power series. It describes the range of values for which the series converges to a finite sum. For the power series \( \sum_{n=0}^{\infty}(-1)^{n} x^{n} \), the interval of convergence is \(-1 < x < 1\). That means within this range, the series will sum to a finite value.

When a power series is differentiated, the interval of convergence usually stays the same. However, one must be cautious about the endpoints of the interval. For the derived series \( \sum_{n=1}^{\infty}(n)(-1)^{n}x^{n-1} \), the same interval \(-1
The endpoints of the interval, \( x = -1 \) and \( x = 1 \), need special attention. Testing these can reveal if the series actually manages to converge at the boundary points. In our case, testing revealed non-convergence at both endpoints.

Here’s a simple checklist when identifying intervals of convergence:

• Evaluate the basic interval from the original series.
• Apply the same interval to any resulting derived series.
• Assess convergence at each endpoint.

Derivative of a Series

Taking the derivative of a power series follows a consistent pattern. Consider the power series \( \sum_{n=0}^{\infty}(-1)^{n} x^{n} \), derived from the function \( f(x) = \frac{1}{x+1} \). You can think of each term individually when differentiating.

The rule here is straightforward: differentiate each term in the series. The exponent of \( x \) decreases by one, and we multiply by the current exponent value \( n \). This transforms the series as follows: \( \sum_{n=1}^{\infty}(n)(-1)^{n}x^{n-1} \).

Here are some quick tips for differentiating power series:

• Remember to shift the index if needed (in this case, starting from \( n=1\) rather than \( n=0\)).
• Decrease the power of \( x \) and multiply by the original power (or index number \( n \)).
• Don't forget that differentiating a series can change the way terms are defined, but typically maintains the original convergence interval.

Convergence Tests

Convergence tests are mathematical tools that help verify if terms in a series converge to a sum. Applying these tests is crucial, especially at the endpoints of convergence intervals.

For a given power series, like \( \frac{1}{1+x} = \sum_{n=0}^{\infty}(-1)^{n}x^{n} \), we naturally use these tests to check if the behavior at \( x = -1 \) and \( x = 1 \) fits within our convergence range.

Common tests include:

• Ratio Test: This checks the absolute limit of successive terms. If the limit is less than one, the series converges absolutely. Not particularly applicable for endpoints here.
• Alternating Series Test: Useful for series that alternate in sign. It ensures that the terms are decreasing in absolute value for convergence.
• Comparison Test: Compare with a known convergent series to establish convergence by similarity.

In our example, checking convergence at the endpoints \( x=-1 \) and \( x=1 \) through simple observation showed divergence. This confirmed that our series remains non-convergent at these boundary conditions.