Find the Interval of Convergence of this Power Series: ∑(x^2n/(2^nn^2))

In summary, the given conversation discusses the use of the ratio test to find the interval of convergence for a power series. It is determined that the interval of convergence is between -√2 and √2, and the endpoints are investigated using the preliminary test. While x = 2 is not one of the endpoints, the answer states that it should be included in the interval.
  • #1
Fernando Rios
96
10
Homework Statement
Find the interval of convergence of each of the following power series; be sure to investigate
the endpoints of the interval in each case.
Relevant Equations
∑(x^(2n)/((2^n)(n^2)))
∑(x2n/(2nn2))

We use the ratio test:
ρn = |(x2n2/(2(n+1)2)|

ρ = |x2/2|

ρ < 1

|x2| < 2

|x| = √(2)

We investigate the endpoints:
x = 2:
∑(4n/(2nn2) = ∑(2n/n2))

We use the preliminary test:
limn→∞ 2n/n2 = ∞

Since the numerator is greater than the denominator. Therefore, x = 2 shouldn't be included. However, then answer says it should be included.
 
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  • #2
Fernando Rios said:
Homework Statement:: Find the interval of convergence of each of the following power series; be sure to investigate
the endpoints of the interval in each case.
Relevant Equations:: ∑(x^(2n)/((2^n)(n^2)))

∑(x2n/(2nn2))

We use the ratio test:
ρn = |(x2n2/(2(n+1)2)|

ρ = |x2/2|

ρ < 1

|x2| < 2

|x| = √(2)

We investigate the endpoints:
x = 2:
This isn't one of the endpoints -- they are ##-\sqrt 2## and ##\sqrt 2##.
Fernando Rios said:
∑(4n/(2nn2) = ∑(2n/n2))

We use the preliminary test:
limn→∞ 2n/n2 = ∞

Since the numerator is greater than the denominator. Therefore, x = 2 shouldn't be included. However, then answer says it should be included.
 
  • #3
Mark44 said:
This isn't one of the endpoints -- they are ##-\sqrt 2## and ##\sqrt 2##.
Thank your for your answer.
 

What is a power series?

A power series is an infinite series of the form ∑(anxn) where an represents the coefficients and x is a variable.

How do you determine the interval of convergence for a power series?

The interval of convergence for a power series is determined by finding the values of x for which the series converges. This can be done by using various convergence tests such as the ratio test or the root test.

What is the ratio test and how is it used to find the interval of convergence?

The ratio test is a convergence test used to determine whether a series converges or diverges. To use the ratio test to find the interval of convergence for a power series, you would take the limit of the absolute value of the ratio of consecutive terms. If the limit is less than 1, the series converges. If the limit is greater than 1, the series diverges. If the limit is equal to 1, further tests are needed to determine convergence.

Can the interval of convergence include the endpoints?

Yes, the interval of convergence can include one or both of the endpoints. This will depend on whether the series converges at the endpoints or not.

How do you use the ratio test to find the interval of convergence for the given power series: ∑(x^2n/(2^nn^2))?

First, we take the limit of the absolute value of the ratio of consecutive terms: limn→∞|an+1/an| = limn→∞|((x^2)^n+1/(2^(n+1)(n+1)^2)) * (2^nn^2)/x^2n|

Simplifying, we get: limn→∞|x^2/(2(n+1)) * (1/(n+1)^2) * (2n^2)/x^2n| = limn→∞|x^2/n+1 * 2n^2 * n^2/x^2n|

Next, we can rewrite x^2 as (x^2)1/n and x^2n as (x^2)n, giving us: limn→∞|x^2/n+1 * 2n^2 * n^2/(x^2)n|

Finally, we can see that as n approaches infinity, x^2/n+1 and n^2/(x^2)n both approach 0, leaving us with the final limit: limn→∞|2n^2 * 0| = 0

Since the limit is less than 1, the given power series converges for all values of x. Therefore, the interval of convergence is (-∞, ∞).

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