# investigate the convergence at the radius of convergence

• March 25th 2011, 07:09 PM
wopashui
investigate the convergence at the radius of convergence
$\displaystyle \sum_{n = 0}^{\infty}$ $\displaystyle \frac{1.5.9...(4n+1)}{3.5.7...(2n+3)}x^n$

so i have calculated R=+or-1/6, and when R=1/6, the series reduced to $\displaystyle \frac{1}{3.7.11.15...(4n+3)} (\displaystyle \frac{1}{6})^n$, now how do we test if this series convergent or not?

The second one

$\displaystyle \sum_{n = 4}^{\infty} \displaystyle \frac{(n-1)!(n-4)!}{[7.10.13...(3n-2)]^2}x^{2n}$

for this one, I have got R=+or- 9/7, and the series reduced to $\displaystyle \frac{(n-1)!(n-4)!}{[7.10.13...(3n-2)]^2}(9/7)^{2n}$, what is the next step?
• March 25th 2011, 09:02 PM
matheagle
how did you get R=1/6?

I thought it was 1/2 as well, but I did this really quick.
I wanted to see the poster's work.
• March 25th 2011, 09:19 PM
Prove It
1. Your series is $\displaystyle \sum_{n = 0}^{\infty}\left[\frac{1\cdot 5\cdot 9 \cdot \dots \cdot(4n+1)}{3\cdot 5 \cdot 7 \cdot \dots \cdot (2n + 3)}x^n\right]$.

Using the ratio test

$\displaystyle \lim_{n \to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty}\left|\frac{\frac{1\cdot 5\cdot 9 \cdot \dots \cdot(4n+1)[4(n+1)+1]}{3\cdot 5 \cdot 7 \cdot \dots \cdot (2n + 3)[2(n+1)+3]}x^{n+1}}{\frac{1\cdot 5\cdot 9 \cdot \dots \cdot(4n+1)}{3\cdot 5 \cdot 7 \cdot \dots \cdot (2n + 3)}x^n}\right|$

$\displaystyle = \lim_{n\to \infty}\left|\left[\frac{4(n+1)+1}{2(n+1)+3}\right]x\right|$

$\displaystyle = \lim_{n \to \infty}\left|\left(\frac{4n + 5}{2n + 5}\right)x\right|$

$\displaystyle = \lim_{n \to \infty}\left|\left(\frac{4 + \frac{5}{n}}{2 + \frac{5}{n}}\right)x\right|$

$\displaystyle = \left|\left(\frac{4}{2}\right)x\right|$

$\displaystyle = \left|2x\right|$.

The series is convergent when this limit $\displaystyle <1$. So

$\displaystyle |2x| < 1$

$\displaystyle -1 < 2x < 1$

$\displaystyle -\frac{1}{2} < x < \frac{1}{2}$.

So the radius of convergence is $\displaystyle \frac{1}{2}$, not $\displaystyle \frac{1}{6}$...

Now substitute each of the endpoints of this interval of convergence into your original series, and test the convergence of those series...
• March 26th 2011, 09:54 AM
wopashui
Quote:

Originally Posted by Prove It
1. Your series is $\displaystyle \sum_{n = 0}^{\infty}\left[\frac{1\cdot 5\cdot 9 \cdot \dots \cdot(4n+1)}{3\cdot 5 \cdot 7 \cdot \dots \cdot (2n + 3)}x^n\right]$.

Using the ratio test

$\displaystyle \lim_{n \to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty}\left|\frac{\frac{1\cdot 5\cdot 9 \cdot \dots \cdot(4n+1)[4(n+1)+1]}{3\cdot 5 \cdot 7 \cdot \dots \cdot (2n + 3)[2(n+1)+3]}x^{n+1}}{\frac{1\cdot 5\cdot 9 \cdot \dots \cdot(4n+1)}{3\cdot 5 \cdot 7 \cdot \dots \cdot (2n + 3)}x^n}\right|$

$\displaystyle = \lim_{n\to \infty}\left|\left[\frac{4(n+1)+1}{2(n+1)+3}\right]x\right|$

$\displaystyle = \lim_{n \to \infty}\left|\left(\frac{4n + 5}{2n + 5}\right)x\right|$

$\displaystyle = \lim_{n \to \infty}\left|\left(\frac{4 + \frac{5}{n}}{2 + \frac{5}{n}}\right)x\right|$

$\displaystyle = \left|\left(\frac{4}{2}\right)x\right|$

$\displaystyle = \left|2x\right|$.

The series is convergent when this limit $\displaystyle <1$. So

$\displaystyle |2x| < 1$

$\displaystyle -1 < 2x < 1$

$\displaystyle -\frac{1}{2} < x < \frac{1}{2}$.

So the radius of convergence is $\displaystyle \frac{1}{2}$, not $\displaystyle \frac{1}{6}$...

Now substitute each of the endpoints of this interval of convergence into your original series, and test the convergence of those series...

should we not just ignore the term x^n when we do the ration test, this is how I do it

$|\displaystyle \frac{a_{n+1}}{a_n}| = \displaystyle \frac{5.9.13.17...(4n+5)}{5.7.9.11...(2n+5)} \displaystyle \frac{3.5.7...(2n+3)}{1.5.9...(4n+1)}$
reduced to $\displaystyle \frac{3(4n+5)}{2n+5}= \displaystyle \frac{12+15/n}{2+5/n}$ which approach to 6 as n goes to infinity, so R is 1/6
• March 26th 2011, 06:38 PM
Prove It
Absolutely not! The changing $\displaystyle x$ terms will change the value of the series.

The whole point in using the ratio test is to find the values of $\displaystyle x$ for which the series is going to converge. If you leave them out, then you're only determining the convergence of the series when $\displaystyle x=1$.
• March 27th 2011, 11:31 AM
wopashui
Quote:

Originally Posted by Prove It
Absolutely not! The changing $\displaystyle x$ terms will change the value of the series.

The whole point in using the ratio test is to find the values of $\displaystyle x$ for which the series is going to converge. If you leave them out, then you're only determining the convergence of the series when $\displaystyle x=1$.

well. we do have a theorem: If $lim_{n-->infinity} |\displaystyle \frac{a_{n+1}}{a_n}|=L>0$, then $\displaystyle \sum$ $a_nx^n$ has a radius convergence $R=1/L$. If L=0, then R=infinity and if L-->infinity as n-->infinity then R=0.
• March 27th 2011, 06:30 PM
Prove It
You're not listening to me.

The $\displaystyle x^n$ is part of $\displaystyle a_n$. If you're going to use the ratio test, then you need to use ALL of $\displaystyle a_{n + 1}$ and $\displaystyle a_n$. In fact, the $\displaystyle x$ term is the most important part, because when you have evaluated what $\displaystyle \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|$ is, you will get an expression for $\displaystyle x$. THIS IS THE IMPORTANT PART, because you know that the series will converge when this limit is $\displaystyle <1$, which means you can solve to find the values of $\displaystyle x$ for which the series will converge. THIS abosolute value is the radius of convergence, and the resulting interval is the interval of convergence.
• March 27th 2011, 07:42 PM
matheagle
i am so listening