Question

Find the interval of convergence, including end-point tests: $$\sum_{n=1}^{\infty} \frac{(-1)^{n} x^{2 n-1}}{2 n-1}$$

Short answer

The interval of convergence is \([-1, 1]\).

Step-by-step solution

- Identify the general term and ratio test

The given series is \(\sum_{n=1}^{\infty} \frac{(-1)^{n} x^{2 n-1}}{2 n-1}\). Let \(a_n = \frac{(-1)^n x^{2n-1}}{2n-1}\). We need to use the ratio test to find the radius of convergence.

- Apply the Ratio Test

The ratio test formula is \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\). First, calculate \(a_{n+1} = \frac{(-1)^{n+1} x^{2(n+1)-1}}{2(n+1)-1} = \frac{(-1)^{n+1} x^{2n+1}}{2n+1}\). Then \( \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{\frac{(-1)^{n+1} x^{2n+1}}{2n+1}}{\frac{(-1)^n x^{2n-1}}{2n-1}} \right| = \left| (-1) \cdot x^2 \cdot \frac{2n-1}{2n+1} \right| = x^2 \cdot \frac{2n-1}{2n+1} \)

- Take the Limit

Simplify the limit: \(\lim_{n \to \infty} x^2 \cdot \frac{2n-1}{2n+1} = x^2 \cdot \lim_{n \to \infty} \frac{2n-1}{2n+1} = x^2 \cdot 1 = x^2\). For the Ratio Test, we require this limit to be less than 1 for convergence: \(x^2 < 1\), hence \(-1 < x < 1\).

- Test End Points

Test \(x = -1\): The series becomes \(\sum_{n=1}^{\infty} \frac{(-1)^n (-1)^{2n-1}}{2n-1} = \sum_{n=1}^{\infty} \frac{(-1)^n (-1)^{2n} (-1)}{2n-1} = \sum_{n=1}^{\infty} \frac{-(-1)^n}{2n-1}\), which simplifies to \(-\sum_{n=1}^{\infty} \frac{(-1)^n}{2n-1}\). This is the alternating harmonic series and converges. Next, test \(x = 1\): The series becomes \(\sum_{n=1}^{\infty} \frac{(-1)^n (1)^{2n-1}}{2n-1} = \sum_{n=1}^{\infty} \frac{(-1)^n}{2n-1}\), which is also the alternating harmonic series and converges.

- Conclude Interval of Convergence

The interval must be \(-1 \leq x \leq 1\). Both endpoints turn the series into the alternating harmonic series, which converges.

Key concepts

ratio test

To determine whether an infinite series converges, the ratio test is a powerful tool. The ratio test revolves around calculating the limit of the absolute value of the ratio of successive terms. Specifically, the formula is \[ \text{Ratio} = \lim_{{n \to \infty}} \left| \frac{{a_{{n+1}}}}{{a_n}} \right| \].
In practice, you compare this limit to 1.
If the result is less than 1, the series converges.
If it’s greater than 1, the series diverges.
If it equal to 1, the test is inconclusive and you need to use another test.

For our given series \(\sum_{{n=1}}^{{\infty}} \frac{{(-1)^n \cdot x^{{2n-1}}}}{{2n-1}}\)\, the formula requires us to find \[ \frac{{a_{{n+1}}}}{{a_n}} = \frac{{(-1)^{{n+1}} \cdot x^{{2(n+1)-1}}}}{{2(n+1)-1}} \cdot \frac{{2n-1}}{{(-1)^n \cdot x^{{2n-1}}}} \].
By simplifying, you obtain \[ \left|(-1) \cdot x^2 \cdot \frac{{2n-1}}{{2n+1}}\right| = x^2 \cdot \frac{{2n-1}}{{2n+1}} \].
Then the limit is \[ x^2 \cdot \lim_{{n \to \infty}} \frac{{2n-1}}{{2n+1}} = x^2 \cdot 1 = x^2 \].
We want this to be less than 1 for convergence, so \[-1 < x < 1\].

alternating series

An alternating series is one in which the terms switch signs between positive and negative as they progress.
In our example, the terms \(\frac{{(-1)^n \cdot x^{{2n-1}}}}{{2n-1}}\)\ alternate between positive and negative.

The Alternating Series Test (Leibniz's criterion) can determine convergence. According to this test, the series \sum (-1)^n b_n \ converges if three conditions are met:

• The terms (b_n) are monotonically decreasing (each term is smaller or equal to the previous one).
• Each term (b_n) is positive.
• The limit of b_n as n approaches infinity is zero: \lim_{{n \to \infty}}b_n = 0.

In our series at endpoints \(-1\) and \(1\) we get \(-1\) and \(1\) turns the series into the alternating harmonic series \(\frac{{(-1)^n}}{{2n-1}}\), which meets all three conditions and thus converges.

harmonic series

The harmonic series is given by \(\sum_{{n=1}}^\{\infty}} \frac{{1}}{n}\), which diverges.
Even though the series is divergent, variations can converge, especially when involving alternating terms.
The series \(- \sum_{{n=1}}^\{\infty}} \frac{{(-1)^n}}{{2n-1}}\) is an alternating harmonic series, which does converge.

The alternating harmonic series converges due to the alternating signs and the terms consistently shrinking (which we discussed in the alternating series section).
Thus, even though the harmonic series \(\sum_{{n=1}}^\{\infty}} \frac{{1}}{n}\) does not converge, our specific series does converge when alternating.

convergence tests

There are several tests used to determine if a series converges or not, and it’s crucial to select the appropriate one:

• Ratio Test: Compares the ratio of successive terms. Converges if \(\lim_{{n \to \infty}} \frac{{a_{{n+1}}}}{{a_n}} < 1\).
• Alternating Series Test: Looks at series where terms alternate in sign.
• Integral Test: Involves integrating the function related to the series. If the integral converges, so does the series.
• Comparison Test: Compares to a series that is already known. If a series smaller than a known convergent series \text{converges}, the new series also converges. Same with divergence.

For our example, the ratio test helped find the radius of convergence. The alternating series test evaluated convergence at the endpoints, confirming that both converged.