What Are the Radii and Intervals of Convergence for These Series?

[KarmaSquared]

Radius of convergence, interval of convergence

Homework Statement

Find the radius of convergence and the interval of convergence of the following series.

a) [tex]\sum_{n=0}^\infty \frac{x^n}{(n^2)+1}[/tex]

c) [tex]\sum_{n=2}^\infty \frac{x^n}{ln(n)}[/tex]

e) [tex]\sum_{n=1}^\infty \frac{n!x^n}{n^2}[/tex]

f) [tex]\sum_{n=1}^\infty \frac{n!2^nx^n}{n^n}[/tex]

Homework Equations

I think I use the Ratio test:

[tex]\lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n} = L < 1[/tex]

The Attempt at a Solution

a) Let [tex]{a_n} = \frac{x^n}{(n^2)+1}[/tex]
So:

[tex]\lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n} = \lim_{n\rightarrow\infty} \frac{x^(n+1)}{((n+1)^2)+1} * \frac{n^2+1}{x^n} = \lim_{n\rightarrow\infty} \frac{x*n^2+1}{n^2+2n+2} = \lim_{n\rightarrow\infty} x[/tex]
Radius: 1, Interval of convergence (-1, 1)

c) Let [tex]{a_n} = \frac{x^n}{ln(n)}[/tex]
So:

[tex]\lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n} = \lim_{n\rightarrow\infty} \frac{x^(n+1)}{ln(n+1)} * \frac{ln(n)}{x^n} = \lim_{n\rightarrow\infty} \frac{x*ln(n)}{ln(n+1)} = \lim_{n\rightarrow\infty} x[/tex]
Radius: 1, Interval of convergence (-1, 1)

e) Let [tex]{a_n} = \frac{n!x^n}{n^2}[/tex]
So:

[tex]\lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n} = \lim_{n\rightarrow\infty} \frac{(n+1)!x^(n+1)}{(n+1)^2} * \frac{n^2}{n!x^n} = \lim_{n\rightarrow\infty} \frac{n^2x}{n+1} = \lim_{n\rightarrow\infty} x[/tex]
Radius: 1, Interval of convergence (-1, 1)

f) Let [tex]{a_n} = \frac{n!2^nx^n}{n^n}[/tex]
So:

[tex]\lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n} = \lim_{n\rightarrow\infty} \frac{n+1!2^(n+1)x^(n+1)}{(n+1)^(n+1)} * \frac{n^n}{n!2^nx^n} = \lim_{n\rightarrow\infty} 2x * (\frac{n}{n+1})^n = \lim_{n\rightarrow\infty} 2x[/tex]
Radius: 1/2, Interval of convergence (-0.5, 0.5)

 

[dynamicsolo]

The radii for three of these seem alright; you may wish to check your ratio in part (e). A number of these series converge for at least one of the endpoints of their intervals of convergence.

 

[KarmaSquared]

The radii for three of these seem alright; you may wish to check your ratio in part (e). A number of these series converge for at least one of the endpoints of their intervals of convergence.

I thought it seemed weird that three of the radii were 1, thanks for checking them over :)

May I inquire which part of (e) you think I might have stumbled? Is it from:
[tex]\lim_{n\rightarrow\infty} \frac{(n+1)!x^(n+1)}{(n+1)^2} * \frac{n^2}{n!x^n} = \lim_{n\rightarrow\infty} \frac{n^2x}{n+1}[/tex]

or

[tex]\lim_{n\rightarrow\infty} \frac{n^2x}{n+1} = \lim_{n\rightarrow\infty} x[/tex]

Will check the endpoints in a bit, thanks again.

 

[cristo]

I thought it seemed weird that three of the radii were 1, thanks for checking them over :)

May I inquire which part of (e) you think I might have stumbled? Is it from:
[tex]\lim_{n\rightarrow\infty} \frac{(n+1)!x^{n+1}}{(n+1)^2} * \frac{n^2}{n!x^n} = \lim_{n\rightarrow\infty} \frac{n^2x}{n+1}[/tex]

or

[tex]\lim_{n\rightarrow\infty} \frac{n^2x}{n+1} = \lim_{n\rightarrow\infty} x[/tex]

Will check the endpoints in a bit, thanks again.

Just a note on LaTex: if you want exponents to show up you need to enclose them in curly brackets; i.e. x^{n+1}.

 

[KarmaSquared]

Oh thanks. I'm still working my way around with LaTeX.

 

[dynamicsolo]

This is not an equation:

[tex]\lim_{n\rightarrow\infty} \frac{n^2x}{n+1} = \lim_{n\rightarrow\infty} x[/tex]

The left-hand side is correct. I believe you should find the radius of convergence for this series to be R = 0. (In fact, the original general term for this series should make one suspicious...)