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?