1. ## 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?

2. 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.

3. 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...

4. 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

5. 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 $\displaystyle x=1$.

6. 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.

7. 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.

8. i am so listening