Radius and Interval of Convergence

[Hip2dagame]

Homework Statement

Find the radius and interval of convergence for the two series:

1) [(n+1)/n]^n * (x^n), series starting at n=1.

2) ln(n)(x^n), series starting at n=1.

Homework Equations

You're usually supposed to root or ratio your way through these.

The Attempt at a Solution

1) First, combining the whole thing and putting to the nth power:
[( (n+1) x) / n] ^ n

then using the root test yields

((n+1)/n) * x

but (n+1)/n doesn't converge. The book still gives R = 1 and I = -1 < x < 1, though... how?

2) Uhhh not too sure... I just ratio tested it and got:

(ln(n+1) / ln(n)) * x

and the ln(n+1) / ln(n) doesn't converge. What do?

Thanks.

 

[LCKurtz]

Homework Statement

Find the radius and interval of convergence for the two series:

1) [(n+1)/n]^n * (x^n), series starting at n=1.

2) ln(n)(x^n), series starting at n=1.

Homework Equations

You're usually supposed to root or ratio your way through these.

The Attempt at a Solution

1) First, combining the whole thing and putting to the nth power:
[( (n+1) x) / n] ^ n

then using the root test yields

((n+1)/n) * x

but (n+1)/n doesn't converge.

Are you sure about that? Write it as$$
1 +\frac 1 n$$

The book still gives R = 1 and I = -1 < x < 1, though... how?

2) Uhhh not too sure... I just ratio tested it and got:

(ln(n+1) / ln(n)) * x

and the ln(n+1) / ln(n) doesn't converge. What do?

Thanks.

For the second one think about L'Hospital's rule.

 

[Hip2dagame]

Ok so giving this a second try, for the first one,
using the root test, we end up with

(1 + (1/n)) * x,

so the limit is absolute value of x, which must be less than 1 to converge, so we set the equation

-1< x< 1

plugging -1 into x we get , (-1)^n * (1+(1/n))^n , a diverging alternating series
plugging 1 into x we get (1)^n (1+ (1/n))^n, which diverges,

so radius of convergence (R) = 1, and interval of con (I) = -1<x<1, open interval.

For number 2,
using LHospital's rule, limit is once again x, abs val of x is less than 1 for convergence so
-1 < x < 1 for convergence,

then plugging -1 in for x, we get (-1)^n * ln(n) which diverges,
plugging in 1 for x, we get (1)^n * ln(n), which diverges,

so R = 1, and I is the open interval -1 < x < 1.

Amiriteuguise? thanks for any help.

 

[LCKurtz]

It looks like you have the idea. Just remember that it is the ratio test or root test you are using, and when you say, for example, "then plugging -1 in for x, we get (-1)^n * ln(n) which diverges", what you really want to say is that (-1)^n * ln(n) doesn't converge to zero, hence the series $\sum_{n=1}^\infty (-1)^n\ln n$ diverges.

And who/what is Amiriteuguise? Something to do with this thread?

 

[rezeew]

The radius of convergence for a power series is the distance from the center of the series (in this case, x=0) to the point where the series converges. The interval of convergence is the range of values for x where the series converges.

For the first series, you have correctly applied the root test, but you also need to consider the limit as n approaches infinity. In this case, the limit is equal to x. So for the series to converge, the absolute value of x must be less than 1. This gives us the interval of convergence -1 < x < 1.

For the second series, you have correctly applied the ratio test, but you also need to consider the limit as n approaches infinity. In this case, the limit is equal to x. So for the series to converge, the absolute value of x must be less than 1. This gives us the interval of convergence -1 < x < 1. However, keep in mind that the series may converge or diverge at the endpoints of the interval. You will need to do further analysis to determine if the series converges or diverges at x=1 and x=-1.