Chapter 1: Q.10.9P (page 22)

$\sum_{n = 1}^{\infty} {(- 1)}^{n} n^{3} x^{n}$

Short Answer

Expert verified

The interval of convergence is $(- 1, 1)$

Step by step solution

Given Information

The power series is $\sum_{n = 1}^{\infty} {(- 1)}^{n} n^{3} x^{n}$

Definition of the interval of convergence.

The Interval of convergence is the interval in which the power series is convergent.

Find the interval.

The power series is $\sum_{n = 1}^{\infty} {(- 1)}^{n} n^{3} x^{n}$

Let $p_{n} = |\frac{a_{n - 1}}{a_{n}}|$ .

Substitute the value of the power series in the formula above, the equation becomes as follows,

$ρ_{n} = |\frac{a_{n + 1}}{a_{n}}|$

$= |\frac{(n + 1)^{3} x^{n + 1}}{n^{3} x^{n}}|$

$= |\frac{(n + 1)^{3} x}{n^{3}}|$

Apply limits in the above equation,

$ρ \lim_{n \to \infty} = |{(1 + \frac{1}{n})}^{3} x|$

$= |x|$

The power series is convergent for $ρ < 1$ ,

Hence the given power series is convergent for $|x| < 1$ and divergent for $|x| > 1$ .

Now check the ends points,

When $x = 1$ series is $\sum_{n = 1}^{\infty} {(- 1)}^{n} n^{3}$ , which is a divergent alternating series.

When $x = - 1$ series is $\sum_{n = 1}^{\infty} n^{3}$ , which is a divergent series.

Hence the interval of convergence is $(- 1, 1)$ .

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$\sum_{n = 1}^{\infty} {(- 1)}^{n} n^{3} x^{n}$

Short Answer

Step by step solution

Given Information

Definition of the interval of convergence.

Find the interval.

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∑n=1∞(-1)nn3xn

Short Answer

Step by step solution

Given Information

Definition of the interval of convergence.

Find the interval.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Electrostatics

Torque and Rotational Motion

Fluids

Classical Mechanics

Dynamics

Measurements

Study anywhere. Anytime. Across all devices.

$\sum_{n = 1}^{\infty} {(- 1)}^{n} n^{3} x^{n}$