Series Convergence: Understanding the Given Series

The given series is an infinite series: \(x - \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} - \frac{{{x^4}}}{4} + \frac{{{x^5}}}{5} - \ldots \infty\). This is an alternating series where the signs of the terms switch. We can recognize this specific series as the Maclaurin series expansion for the natural logarithm function, \(\ln(1+x)\).

The general term of the series can be written as \(a_n = (-1)^{n-1} \frac{x^n}{n}\) for \(n \ge 1\).

Convergence Analysis: Applying the Ratio Test

To find the interval of convergence for this series, we use the Ratio Test. The Ratio Test states that an infinite series \(\sum a_n\) converges if the limit of the absolute value of the ratio of consecutive terms is less than 1, i.e., \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1\).

Let's calculate the ratio \(\left| \frac{a_{n+1}}{a_n} \right|\):

\[ \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(-1)^{(n+1)-1} \frac{x^{n+1}}{n+1}}{(-1)^{n-1} \frac{x^n}{n}} \right| \] \[ = \left| \frac{(-1)^n x^{n+1}}{n+1} \cdot \frac{n}{(-1)^{n-1} x^n} \right| \] \[ = \left| \frac{(-1)^n}{(-1)^{n-1}} \cdot \frac{x^{n+1}}{x^n} \cdot \frac{n}{n+1} \right| \] \[ = \left| (-1) \cdot x \cdot \frac{n}{n+1} \right| \] \[ = |x| \frac{n}{n+1} \]

Now, we take the limit as \(n\) approaches infinity:

\[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left( |x| \frac{n}{n+1} \right) \] \[ = |x| \lim_{n \to \infty} \left( \frac{1}{1 + 1/n} \right) \] \[ = |x| \cdot 1 = |x| \]

For convergence, according to the Ratio Test, we must have \(|x| < 1\). This means the series converges for \(-1 < x < 1\). The Ratio Test is inconclusive when \(|x| = 1\), so we need to check the endpoints separately.

Endpoint Convergence: Examining x = 1

We substitute \(x = 1\) into the original series:

\[ 1 - \frac{1^2}{2} + \frac{1^3}{3} - \frac{1^4}{4} + \frac{1^5}{5} - \ldots \]

This simplifies to the Alternating Harmonic Series:

\[ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \ldots \]

We can use the Alternating Series Test to check its convergence. For an alternating series \(\sum (-1)^{n-1} b_n\) to converge, two conditions must be met:

• The sequence \(b_n = \frac{1}{n}\) must be positive and decreasing. For \(n \ge 1\), \(b_n > 0\) and \(b_{n+1} = \frac{1}{n+1} \le \frac{1}{n} = b_n\). So, it's decreasing.
• The limit of \(b_n\) as \(n\) approaches infinity must be zero. \(\lim_{n \to \infty} \frac{1}{n} = 0\).

Since both conditions are satisfied, the series converges at \(x = 1\).

Endpoint Convergence: Examining x = -1

Next, we substitute \(x = -1\) into the original series:

\[ (-1) - \frac{(-1)^2}{2} + \frac{(-1)^3}{3} - \frac{(-1)^4}{4} + \frac{(-1)^5}{5} - \ldots \]

This simplifies to:

\[ -1 - \frac{1}{2} - \frac{1}{3} - \frac{1}{4} - \frac{1}{5} - \ldots \]

We can factor out -1:

\[ = -\left( 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \ldots \right) \]

The series inside the parenthesis is the Harmonic Series, \(\sum_{n=1}^{\infty} \frac{1}{n}\). The Harmonic Series is a known divergent series.

Since the series \(1 + \frac{1}{2} + \frac{1}{3} + \ldots\) diverges, its negative also diverges. Therefore, the series diverges at \(x = -1\).

Series Convergence: Final Interval

Combining the results from the Ratio Test and the endpoint checks:

• The series converges for \(-1 < x < 1\) based on the Ratio Test.
• The series converges at \(x = 1\).
• The series diverges at \(x = -1\).

Thus, the given series converges for all values of \(x\) such that \(-1 < x \le 1\).

Comparing this with the provided options, the correct interval for series convergence is \(-1 < x \le 1\).