Intervals of Convergence
In Exercises 1–36, (a) find the series’ radius and interval of convergence. For what values of x does the series converge (b) absolutely, (c) conditionally?
∑ (from n = 1 to ∞) [ (x − 1)ⁿ / (n³ 3ⁿ) ]

Verified step by step guidance

Identify the given power series: \(\sum_{n=1}^{\infty} \frac{(x - 1)^n}{n^3 3^n}\).

To find the radius of convergence, apply the Root Test or Ratio Test. Here, the Ratio Test is convenient. Consider the general term \(a_n = \frac{(x - 1)^n}{n^3 3^n}\) and compute the limit \(L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\).

Calculate the ratio: \(\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(x - 1)^{n+1}}{(n+1)^3 3^{n+1}} \cdot \frac{n^3 3^n}{(x - 1)^n} \right| = \left| \frac{x - 1}{3} \right| \cdot \frac{n^3}{(n+1)^3}\).

Evaluate the limit as \(n \to \infty\): \(L = \left| \frac{x - 1}{3} \right| \cdot \lim_{n \to \infty} \frac{n^3}{(n+1)^3} = \left| \frac{x - 1}{3} \right|\).

The Ratio Test states the series converges if \(L < 1\), so the radius of convergence \(R\) satisfies \(\left| x - 1 \right| < 3\). This gives the interval \((1 - 3, 1 + 3)\) or \((-2, 4)\). Next, check convergence at the endpoints \(x = -2\) and \(x = 4\) by substituting back into the series and analyzing absolute and conditional convergence.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

15m

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Radius and Interval of Convergence

The radius of convergence is the distance from the center of a power series within which the series converges. The interval of convergence includes all x-values for which the series converges, possibly including endpoints. Finding these involves applying tests like the Ratio or Root Test to determine where the series converges absolutely.

Absolute Convergence

A series converges absolutely if the series of absolute values converges. This means ∑|a_n| converges, ensuring the original series converges regardless of term signs. Absolute convergence implies stronger convergence and often simplifies analysis of power series.

Conditional Convergence

Conditional convergence occurs when a series converges, but does not converge absolutely. This means the series ∑a_n converges, but ∑|a_n| diverges. Identifying conditional convergence often requires testing endpoints of the interval of convergence separately.

Recommended video:

07:51

Choosing a Convergence Test

Related Practice

Textbook Question

Determining Convergence or Divergence

Which of the series in Exercises 13–46 converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series’ convergence or divergence.)

∑ (from n=1 to ∞) eⁿ / (1 + e²ⁿ)

views

Textbook Question

Finding Terms of a Sequence

Each of Exercises 1–6 gives a formula for the nth term aₙ of a sequence {aₙ}. Find the values of a₁, a₂, a₃, and a₄.

aₙ = 1 / n!

views

Textbook Question

Absolute and Conditional Convergence

Which of the series in Exercises 15–48 converge absolutely, which converge, and which diverge? Give reasons for your answers.

∑ (from n = 1 to ∞) [(-1)ⁿ (√(n² + n) − n)]

views