# Possible Values of x for Convergence of Power Series

• Fernando Rios
In summary, the conversation discusses transforming a series into a power series by a change of variable. The ratio test is used to find the possible values of "x" for the series, but the result is incorrect due to an error in substituting ∞ into an arithmetic expression. The correct answer is |x| < √(5)/2.
Fernando Rios
Homework Statement
The following series are not power series, but you can transform each one into a power
series by a change of variable and so find out where it converges.
Relevant Equations
Σ((√(x^2+1))^n 2^n/(3^n+n^3))
We transform the series into a power series by a change of variable:
y = √(x2+1)

We have the following after substituting:
∑(2nyn/(3n+n3))

We use the ratio test:
ρn = |(2n+1yn+1/(3n+1+(n+1)3)/(2nyn/(3n+n3)| = |(3n+n3)2y/(3n+1+(n+1)3)|

ρ = |(3+∞3)2y/(3∞+1+(∞+1)3)| = |2y|

|2y| < 1

|y| = 1/2

-1/2 < y < 1/2

We find the possible values of "x":
-1/2 < √(x2+1) < 1/2

This inequality has no solution, so I conclude the series doesn't converge for any "x", but the book says the answer is |x| < √(5)/2. Could you please tell me what am I doing wrong?

Fernando Rios said:
Homework Statement:: The following series are not power series, but you can transform each one into a power
series by a change of variable and so find out where it converges.
Relevant Equations:: Σ((√(x^2+1))^n 2^n/(3^n+n^3))

We transform the series into a power series by a change of variable:
y = √(x2+1)

We have the following after substituting:
∑(2nyn/(3n+n3))

We use the ratio test:
ρn = |(2n+1yn+1/(3n+1+(n+1)3)/(2nyn/(3n+n3)| = |(3n+n3)2y/(3n+1+(n+1)3)|

ρ = |(3+∞3)2y/(3∞+1+(∞+1)3)| = |2y|
No. You should never substitute ##\infty## into an arithmetic expression.

BTW, I get a limit of ##\frac 2 3 |y|##.
Fernando Rios said:
|2y| < 1

|y| = 1/2
-1/2 < y < 1/2
Aside from the error I mentioned above, in your substitution, y will always be nonnegative, since ##y = \sqrt{x^2 + 1}##.
Fernando Rios said:
We find the possible values of "x":
-1/2 < √(x2+1) < 1/2

This inequality has no solution, so I conclude the series doesn't converge for any "x", but the book says the answer is |x| < √(5)/2. Could you please tell me what am I doing wrong?
[/QUOTE]

## What is a power series?

A power series is an infinite series of the form ∑n=0∞ an(x-c)n, where an are constants and x is a variable. It is a type of mathematical function that can be used to represent other functions, such as polynomials or trigonometric functions.

## What is the convergence of a power series?

The convergence of a power series refers to whether the series converges to a finite value or not. If it does, then the series is said to be convergent. If it does not, then the series is said to be divergent. The convergence of a power series depends on the values of x and the coefficients an.

## How do you determine the convergence of a power series?

To determine the convergence of a power series, you can use various tests such as the ratio test, root test, or the comparison test. These tests involve comparing the given series to a known convergent or divergent series. If the given series behaves similarly to the known series, then it has the same convergence properties.

## What is the radius of convergence of a power series?

The radius of convergence of a power series is the distance from the center point (c) to the nearest point where the series converges. It is denoted by R and can be calculated using the ratio test or the root test. The radius of convergence determines the values of x for which the series will converge.

## What happens at the endpoints of the interval of convergence?

The endpoints of the interval of convergence are the values of x at which the series may or may not converge. To determine if the series converges at the endpoints, you can use the endpoint test. If the series converges at the endpoints, then the interval of convergence is equal to the interval of convergence. If the series diverges at one or both endpoints, then the interval of convergence is a half-open interval.

• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
470
• Calculus and Beyond Homework Help
Replies
1
Views
559
• Calculus and Beyond Homework Help
Replies
4
Views
909
• Calculus and Beyond Homework Help
Replies
2
Views
462
• Calculus and Beyond Homework Help
Replies
3
Views
318
• Calculus and Beyond Homework Help
Replies
1
Views
708
• Calculus and Beyond Homework Help
Replies
2
Views
939
• Calculus and Beyond Homework Help
Replies
14
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
895