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Find the convergence set for the given power series. Hint: First find a formula for the nth term; then use the Absolute Ratio Test. $$ \frac{x}{1 \cdot 2}-\frac{x^{2}}{2 \cdot 3}+\frac{x^{3}}{3 \cdot 4}-\frac{x^{4}}{4 \cdot 5}+\frac{x^{5}}{5 \cdot 6}-\cdots $$

Short Answer

Expert verified
The convergence set for the series is \(-1 < x < 1\).

Step by step solution

01

Identify the nth term of the series

The given power series is \( \frac{x}{1 \cdot 2} - \frac{x^2}{2 \cdot 3} + \frac{x^3}{3 \cdot 4} - \frac{x^4}{4 \cdot 5} + \cdots \). Observing the pattern, the general term of the series (nth term) can be identified as \((-1)^{n+1} \frac{x^n}{n(n+1)}\).
02

Apply the Absolute Ratio Test

The Absolute Ratio Test states that a series \( \sum a_n \) converges if \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \). For our series, \( a_n = (-1)^{n+1} \frac{x^n}{n(n+1)} \). The next term is \( a_{n+1} = (-1)^{(n+1)+1} \frac{x^{n+1}}{(n+1)(n+2)} \).
03

Calculate the ratio \( \frac{a_{n+1}}{a_n} \)

Substitute \( a_n \) and \( a_{n+1} \) into the ratio: \[ \frac{a_{n+1}}{a_n} = \frac{(-1)^{n+2} \frac{x^{n+1}}{(n+1)(n+2)}}{(-1)^{n+1} \frac{x^n}{n(n+1)}}. \] Simplify to get: \[ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{x^{n+1} [(-1)^2]}{x^n} \cdot \frac{n(n+1)}{(n+1)(n+2)} \right| = \left| x \cdot \frac{n}{n+2} \right|. \]
04

Evaluate the limit

Find the limit: \[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| x \cdot \frac{n}{n+2} \right| = |x| \cdot \lim_{n \to \infty} \frac{n}{n+2} = |x| \cdot 1 = |x|. \]
05

Determine the convergence set

According to the Absolute Ratio Test, the series converges when the limit \(|x| < 1\). Therefore, the convergence set for the power series is \(-1 < x < 1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

nth term identification
Identifying the nth term of a power series is a crucial step towards understanding the series and determining its behavior. In the given power series, you observe a repeating pattern that helps you derive the nth term. Initially, you see the numerators are powers of \(x\), specifically \(x^1, x^2, x^3\), and so on. This hints that the general term involves \(x^n\). The denominators are products like \(1 \cdot 2, 2 \cdot 3, 3 \cdot 4\). You can see these correspond to \(n(n+1)\). Additionally, the signs alternate starting from a positive first term, which suggests the use of \((-1)^{n+1}\). By piecing these observations together, the nth term can be expressed as \((-1)^{n+1} \frac{x^n}{n(n+1)}\). This expression captures the general behavior of any term in the series, making analysis possible.
absolute ratio test
The Absolute Ratio Test is a powerful tool for determining the convergence of a power series. It requires taking the ratio of successive terms and examining the behavior as \(n\) approaches infinity. For our series, we start with two terms, \(a_n = (-1)^{n+1} \frac{x^n}{n(n+1)}\) and \(a_{n+1} = (-1)^{n+2} \frac{x^{n+1}}{(n+1)(n+2)}\). The task is to find \(\left| \frac{a_{n+1}}{a_n} \right|\).

By substituting these expressions into the ratio, you begin simplifying: the \((-1)\) terms effectively cancel since \((-1)^2 = 1\). You then focus on the powers of \(x\) and the factorial-like terms. This results in a simpler expression: \(\left| x \cdot \frac{n}{n+2} \right|\). The Absolute Ratio Test concludes that if the limit of this expression is less than 1, the series converges.
convergence interval
Understanding the convergence interval for a power series involves determining where the series behaves and sums up meaningfully. Based on the Absolute Ratio Test, this interval is determined by evaluating when \(|x| < 1\).

For our specific series, once you apply and simplify the ratio test, you arrive at the condition \(|x| < 1\). This tells us that for any value of \(x\) within the interval from -1 to 1, the series converges. It's essential because outside this interval, the terms of the series do not sum to a finite value. The interval is open, meaning at \(x = -1\) or \(x = 1\) the result of the series is not included, thus emphasizing the importance of staying within bounds for practical applications.
limit evaluation
Evaluating limits accurately is a key step in analyzing series convergence. In the context of the Absolute Ratio Test, you need to calculate \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\).

For our series, simplify \(\left| x \cdot \frac{n}{n+2} \right|\) further to see the behavior as \(n\) grows large. As \(n \to \infty\), the fraction \(\frac{n}{n+2}\) approaches 1 since the term \(+2\) in the denominator becomes negligible in comparison. Thus, the limit evaluates to \(|x| \cdot 1\), which simplifies to \(|x|\). Convergence is ensured for \(|x| < 1\), meaning our calculated limit helps in confirming the range over which the series sums correctly.

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