i have to determine the radius of convergence of the following power series, state the regions which the series converge uniformly and study the convergence at the boundary of the interval of convergence

1)

$\displaystyle \sum_{n=0}^\infty k! x^n$

2)

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

2. Originally Posted by TS1
i have to determine the radius of convergence of the following power series, state the regions which the series converge uniformly and study the convergence at the boundary of the interval of convergence

1)

$\displaystyle \sum_{n=0}^\infty k! x^n$

2)

$\displaystyle \sum_{n=1}^\infty \frac{(n!)^2}{(2n)!}x^n$
2. Use the ratio test.

$\displaystyle \left|\frac{a_{n + 1}}{a_n}\right| = \left|\frac{[(n + 1)!]^2 (2n)!x^{n + 1}}{(n!)^2 x^n [2(n + 1)]!}\right|$

$\displaystyle = \left|\frac{(n + 1) x}{2(2n + 1)}\right|$.

So $\displaystyle \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty}\left| \frac{(n + 1) x}{2(2n + 1)}\right|$

$\displaystyle = \lim_{n \to \infty}\left[ \left(\frac{n + 1}{2n + 1}\right) \left|\frac{x}{2}\right|\right]$

$\displaystyle = \lim_{n \to \infty}\left[\left(\frac{1}{2}\right)\left|\frac{x}{2}\right|\r ight]$ by L'Hospital's Rule

$\displaystyle = \left|\frac{x}{4}\right|$.

For this series to converge, the limit is $\displaystyle <1$.

So $\displaystyle \left|\frac{x}{4}\right|<1$

$\displaystyle -1 <\frac{x}{4} < 1$

$\displaystyle -4 < x < 4$.

So the radius of convergence is $\displaystyle 4$. I'll leave you to check the end points.

3. what happens to the convergence at the boundary of the interval of convergence?
thank you

4. Originally Posted by TS1
i have to determine the radius of convergence of the following power series, state the regions which the series converge uniformly and study the convergence at the boundary of the interval of convergence

1)

$\displaystyle \sum_{n=0}^\infty k! x^n$
Is that really k! and not n!? If so, the sum can be written as $\displaystyle k!\sum_{n=0}^\infty x^n$ and the radius of convergence is just the radius of convergence of $\displaystyle \sum_{n=0}^\infty x^n$. As Prove It says, the simplest way to find radii of convergence for power series is to use the ratio test:
If $\displaystyle a_n= x^n$, then $\displaystyle a_{n+1}= x^{n+1}$ and $\displaystyle \left|\frac{a_{n+1}}{a_n}\right|= \left|\frac{x^{n+1}}{x^n}\right|= |x|$. Since the ratio test says that will converge uniformly for the limit of the ratio less than one, we must have |x|< 1 and the radius of convergence is 1.
Now, the boundary points are x= -1 and x= 1. If x= 1, that is k(1+ 1+ 1+ ...). Does that converge? At x= -1, it is k(1- 1+ 1- 1+ ....). Does that converge?

If it is actually $\displaystyle \sum_{n=0}^\infty n! x^n$ it's a little different:
$\displaystyle \left|\frac{a_{n+1}}{a_n}\right|= \left|\frac{(n+1)x^{n+1}}{nx^n}\right|= \frac{n+1}{n} |x|$ and you must take the limit, as n goes to infinity, of that.

2)

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

5. Originally Posted by TS1
what happens to the convergence at the boundary of the interval of convergence?
thank you
Since the ratio test fails when the limit is 1, you need to use some other test at the boundaries (i.e. where the limit is 1).

6. for the second question i should use a different test?? i'm stuck on what to do and what my answer implies?

7. Post it has already told you that the radius of convergence is 4. That means that the endpoints of the interval of convergence are x= -4 and x= 4.

For x= 4, that series is $\displaystyle \sum_{n=0}^\infty \frac{(n!)^2}{(2n)!}4^n$ or $\displaystyle 1+ 2+ \frac{4}{3}+ \frac{24}{5}+ \cdot\cdot\cdot$
That's not going to be easy to decide!

For x= -4 that series is $\displaystyle \sum_{n=0}^\infty \frac{(n!)^2}{(2n)!}(-4)^n$. Because that is an 'alternating' series, it will converge as long as the individual terms are eventually decreasing.

8. Originally Posted by HallsofIvy
Post it has already told you that the radius of convergence is 4. That means that the endpoints of the interval of convergence are x= -4 and x= 4.

For x= 4, that series is $\displaystyle \sum_{n=0}^\infty \frac{(n!)^2}{(2n)!}4^n$ or $\displaystyle 1+ 2+ \frac{4}{3}+ \frac{24}{5}+ \cdot\cdot\cdot$
That's not going to be easy to decide!

For x= -4 that series is $\displaystyle \sum_{n=0}^\infty \frac{(n!)^2}{(2n)!}(-4)^n$. Because that is an 'alternating' series, it will converge as long as the individual terms are eventually decreasing.
Use the simple divergence test, that if $\displaystyle \lim_{n \to \infty}a_n \neq 0$ then the series is divergent...

In this case, you have

$\displaystyle \sum_{n = 1}^{\infty}\frac{(n!)^24^n}{(2n)!}$

According to Wolfram, this limit is $\displaystyle \infty$, not $\displaystyle 0$.

So the series is divergent when $\displaystyle x = 4$.

9. thanks, but how would i find the limit a(n)?
$\displaystyle$
$\displaystyle \sum_{n = 1}^{\infty}\frac{(n!)^24^n}{(2n)!}$