# Finding interval of convergance

• Jul 14th 2013, 12:52 PM
minneola24
Finding interval of convergance
How do I find the interval of convergance for this?

I did the ratio test and get .25<x<.25
The book has .5<x<.5

I will do the endpoint test after I see why I'm not getting one half as my interval!Attachment 28828
• Jul 14th 2013, 01:11 PM
emakarov
Re: Finding interval of convergance
Quote:

Originally Posted by minneola24
I will do the endpoint test after I see why I'm not getting one half as my interval!

This is because you first need to represent the series as $\displaystyle \sum_{n=0}^\infty c_nx^n$ for some $\displaystyle c_n$ and not as $\displaystyle \sum_{n=0}^\infty c_nx^{2n}$.
• Jul 14th 2013, 01:29 PM
minneola24
Re: Finding interval of convergance
Ok.

How exactly do I do that, and why?
• Jul 14th 2013, 01:45 PM
Plato
Re: Finding interval of convergance
Quote:

Originally Posted by minneola24
Ok.
How exactly do I do that, and why?

Use the root on $\displaystyle ~\;\;\sqrt[n]{{\left| {\frac{{{2^{2n + 3}}{x^{2n + 2}}}}{{2n + 2}}} \right|}}$
• Jul 14th 2013, 01:46 PM
Soroban
Re: Finding interval of convergance
Hello, minneola24!

Quote:

$\displaystyle \text{Find the interval of convergence: }\;\sum^{\infty}_{n=0}(-1)^n\frac{4^n\cdot8\cdot x^{2n+2}}{2n+2}$

$\displaystyle |R| \;=\;\left|\frac{a_{n+1}}{a_n}\right| \;=\;\left|\frac{4^{n+1}\cdot 8\cdot x^{2n+4}}{2n+4} \cdot \frac{2n+2}{4^n\cdot8\cdot x^{2n+2}}\right|$

. . . . . . . . . .$\displaystyle =\;\left|\frac{4^{n+1}\cdot8}{4^n\cdot8}\cdot \frac{x^{2n+4}}{x^{2n+2}}\cdot \frac{2n+2}{2n+4}\right|$

. . . . . . . . . .$\displaystyle =\;\left|4\cdot x^2\cdot\frac{n+1}{n+2}\right|$

. . . . . . . . . .$\displaystyle =\;\left|4\cdot x^2\cdot\frac{1+\frac{1}{n}}{1+\frac{2}{n}} \right|$

$\displaystyle \lim_{n\to\infty}|R| \;=\;\lim_{n\to\infty}\left|4\cdot x^2\cdot\frac{1+\frac{1}{n}}{1+\frac{2}{n}} \right| \;=\;\left|4\cdot x^2\cdot\frac{1+0}{1+0}\right| \;=\;|4x^2|$

Then: .$\displaystyle |4x^2| \:<\:1 \quad\Rightarrow\quad |x^2| \:<\:\tfrac{1}{4}$

Therefore: .$\displaystyle |x|\:<\:\tfrac{1}{2} \quad\Rightarrow\quad \text{-}\tfrac{1}{2}\:<\:x \:<\:\tfrac{1}{2}$
• Jul 14th 2013, 03:38 PM
emakarov
Re: Finding interval of convergance
Quote:

Originally Posted by minneola24
How exactly do I do that, and why?

I intended to use facts specifically about the radius of convergence of a power series instead of tests that apply to any series.

$\displaystyle \sum_{n=0}^\infty(-1)^n\frac{4^n\cdot8\cdot x^{2n+2}}{2n+2}=$ (set k = n + 1, i.e., n = k - 1)
$\displaystyle \sum_{k=1}^\infty(-1)^{k-1}\frac{4^{k-1}\cdot8\cdot x^{2k}}{2k}=$
$\displaystyle \sum_{n=1}^\infty c_nx^n$ (set n = 2k, i.e., k = n/2)

where

$\displaystyle c_n= \begin{cases}\displaystyle (-1)^{n/2-1}\frac{4^{n/2-1}\cdot8}{n} & n \text{ is even}\\ 0 & n \text{ is odd} \end{cases}$

Now, strictly speaking, we cannot use the formula

$\displaystyle r = \lim_{n\rightarrow\infty} \left| \frac{c_n}{c_{n+1}} \right|$

because some $\displaystyle c_n$ are equal to 0 (but see Soroban's answer on how to use ratio test for arbitrary series), but we can use

$\displaystyle r = \frac{1}{\limsup_{n\rightarrow\infty}\sqrt[n]{|c_n|}}$

Since $\displaystyle \sqrt[n]{n}\to 1$ as $\displaystyle n\to\infty$, $\displaystyle \limsup_{n\to\infty}|c_n|^{1/n}= \lim_{n\to\infty}\left(4^{n/2-1}\right)^{1/n} =4^{1/2}=2$, so r = 1/2. If we did not convert the series to the form $\displaystyle \sum_{n=1}^\infty c_nx^n$ and took the nth root of $\displaystyle \frac{4^n\cdot8}{2n+2}$, we would get r = 1/4.