Question

Find the interval of convergence of each of the following power series; be sure to investigate the endpoints of the interval in each case. $$\sum_{n=1}^{\infty} \frac{n(-x)^{n}}{n^{2}+1}$$

Short answer

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

Step-by-step solution

Ratio Test for Convergence

Apply the Ratio Test to find the radius of convergence (R). For the series \( \sum_{n=1}^{\infty} \frac{n(-x)^{n}}{n^{2}+1}\), consider the general term \(a_n = \frac{n(-x)^n}{n^2 + 1}\). Compute the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\).

Compute the Limiting Ratio

Find \( \frac{a_{n+1}}{a_n} \). This equals \( \frac{(n+1)(-x)^{n+1}}{(n+1)^2 + 1} \div \frac{n(-x)^n}{n^2 + 1}\), which simplifies to \( \frac{(n+1)(-x)(n^2 + 1)}{n((n+1)^2 + 1)}\). Simplify further to get \( \left|(-x) \cdot \frac{n+1}{n} \cdot \frac{n^2 + 1}{(n+1)^2 + 1}\).

Simplify and Apply the Limit

Simplify inside the limit: \( \left| \frac{(n + 1)(n^2 + 1)}{n((n + 1)^2 + 1)} (-x) \right|\). Evaluate the limit as \( n \to \infty\) to find \( L = |x| \lim_{n \to \infty} \frac{(n+1)(n^2 + 1)}{n((n+1)^2 + 1)} = |x| \cdot 1\) since higher-order terms dominate.

Determine the Radius of Convergence

We have \( L = |x|\). According to the Ratio Test, the series converges when \( L < 1\) and diverges when \( L > 1\). Thus the radius of convergence \( R = 1\).

Finding the Interval of Convergence

The radius of convergence tells \( |x| < 1\). This translates to the interval \( -1 < x < 1\).

Test the Endpoints

Investigate convergence at \( x = -1\) and \( x = 1\). Substitute \( x = -1\) into the original series: \( \sum_{n=1}^{\infty} \frac{n(-(-1))^{n}}{n^{2}+1} = \sum_{n=1}^{\infty} \frac{n(-1)^{n+1}}{n^{2}+1}\). The resulting series alternates and converges by the Alternating Series Test.

Endpoint \( x = 1 \)

Substitute \( x = 1\). The series becomes \( \sum_{n=1}^{\infty} \frac{n(-1)^n}{n^2 + 1}\). This is also an alternating series and converges by the Alternating Series Test.

Conclusion

Since the series converges at \( x = -1\) and \( x = 1\), the interval of convergence is \([-1, 1]\).

Key concepts

ratio test

The Ratio Test is a common method to determine the convergence of a series. To apply the Ratio Test, we use the general term of the series, say \(a_n\). For our series, \(a_n = \frac{n(-x)^n}{n^2 + 1}\). The test requires us to compute the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). This ratio helps us determine whether the series converges or diverges.

We simplify the fraction \(\frac{a_{n+1}}{a_n}\) by dividing the \(n+1\) term by the \(n\) term inside the limit. Finally, the absolute value ensures we only consider the magnitude. If the limit \(L < 1\), the series converges; if \(L > 1\), it diverges. If \(L = 1\), the Ratio Test is inconclusive.

radius of convergence

The Radius of Convergence (R) is derived from the Ratio Test. It tells us the range of values where the series converges. From the Ratio Test, the radius of convergence is given by \( |x| < R \). For our series, after applying the Ratio Test, we obtained \( |x| < 1 \).

This means the series converges when \(|x| < 1\), giving us a radius of 1. The radius explains how far from the center (usually at 0) the series converges. If \(R = 1\), it means the power series converges for all \(x\) within distance 1 from the center.

interval of convergence

The Interval of Convergence extends the concept of radius by including endpoints. With \(|x| < 1\), the radius gives us the open interval (-1, 1). Next, we test the series at the endpoints of the interval: \(x = -1\) and \(x = 1\). Plugging these values into the series helps determine if the series converges at these points.

If it converges at both endpoints, the interval becomes closed \([-1, 1]\). Otherwise, if it diverges at one or both endpoints, we adjust the interval accordingly. For our series, testing the endpoints showed it converged at both, leading to an interval of convergence \([-1, 1]\).

alternating series test

The Alternating Series Test (Leibniz Test) is used to confirm convergence of series that alternate in sign, such as those in the form \( (-1)^n b_n \). Two conditions must be met:

• The terms \(b_n\) must decrease in magnitude.
• The limit of the terms \(b_n\) should approach zero as \(n\) approaches infinity.

In our case, the terms alternated through \( (-1)^n \) or \( (-1)^{n+1} \). After substituting at the endpoints, both series \(\sum_{n=1}^\infty \frac{n(-1)^{n+1}}{n^2+1}\) and \(\sum_{n=1}^\infty \frac{n(-1)^n}{n^2+1}\) met these conditions for the Alternating Series Test, confirming convergence at \(x = -1\) and \(x = 1\).