1. Dec 9, 2007

### KarmaSquared

Radius of convergence, interval of convergence

1. The problem statement, all variables and given/known data

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

a) $$\sum_{n=0}^\infty \frac{x^n}{(n^2)+1}$$

c) $$\sum_{n=2}^\infty \frac{x^n}{ln(n)}$$

e) $$\sum_{n=1}^\infty \frac{n!x^n}{n^2}$$

f) $$\sum_{n=1}^\infty \frac{n!2^nx^n}{n^n}$$

2. Relevant equations

I think I use the Ratio test:

$$\lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n} = L < 1$$

3. The attempt at a solution

a) Let $${a_n} = \frac{x^n}{(n^2)+1}$$
So:

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

c) Let $${a_n} = \frac{x^n}{ln(n)}$$
So:

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

e) Let $${a_n} = \frac{n!x^n}{n^2}$$
So:

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

f) Let $${a_n} = \frac{n!2^nx^n}{n^n}$$
So:

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

Last edited: Dec 9, 2007
2. Dec 9, 2007

### 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.

3. Dec 9, 2007

### KarmaSquared

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:
$$\lim_{n\rightarrow\infty} \frac{(n+1)!x^(n+1)}{(n+1)^2} * \frac{n^2}{n!x^n} = \lim_{n\rightarrow\infty} \frac{n^2x}{n+1}$$

or

$$\lim_{n\rightarrow\infty} \frac{n^2x}{n+1} = \lim_{n\rightarrow\infty} x$$

Will check the endpoints in a bit, thanks again.

4. Dec 9, 2007

### cristo

Staff Emeritus
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}.

5. Dec 9, 2007

### KarmaSquared

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

6. Dec 9, 2007

### dynamicsolo

This is not an equation:

$$\lim_{n\rightarrow\infty} \frac{n^2x}{n+1} = \lim_{n\rightarrow\infty} x$$

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...)