# Power Series

• May 12th 2009, 08:46 PM
aaronb
Power Series
I am finding the radius of convergence and interval of convergence.

the sum of n=1 to infinity of x^n/(6^n*n^4)

i used the ratio test on this and found that the limit goes to 1 and the radius of convergence is 6.

when x is -6, i know the series alternates but when i compare b of n to b of n+1 I don't see how b of n is greater than b of n+1 which is one of the stipulations for the alternating series test.
• May 12th 2009, 08:54 PM
TheEmptySet
Quote:

Originally Posted by aaronb
I am finding the radius of convergence and interval of convergence.

the sum of n=1 to infinity of x^n/(6^n*n^4)

i used the ratio test on this and found that the limit goes to 1 and the radius of convergence is 6.

when x is -6, i know the series alternates but when i compare b of n to b of n+1 I don't see how b of n is greater than b of n+1 which is one of the stipulations for the alternating series test.

$a_n=\frac{x^n}{n^4\cdot6^n}$

$a_{n+1}=\frac{x^{n+1}}{(n+1)^4\cdot6^{n+1}}=\frac{ x}{6}\cdot\frac{x^n}{(x+1)^46^n } = 1\cdot \frac{x^n}{(x+1)^46^n } <\frac{x^n}{n^46^n}=a_n$

Remember that $\frac{x}{6}= 1$ at the egde of the radius of convergence
• May 13th 2009, 04:17 AM
Krizalid
root test is faster than ratio test.
• May 13th 2009, 04:57 AM
Soroban
Hello, aaronb!

Interesting . . . I've seen the Ratio Test described in several ways now.
. . And a few of them never form a ratio.

Quote:

Find the radius of convergence and interval of convergence:
. . $\sum^{\infty}_{n=1} \frac{x^n}{6^nn^4}$

$R \:=\:\left|\frac{a_{n+1}}{a_n}\right| \;=\;\left|\frac{x^{n+1}}{6^{n+1}(n+1)^4}\cdot\fra c{6^nn^4}{x^n}\right| \;=\;\left|\frac{6^n}{6^{n_1}}\cdot\frac{n^4}{(n+1 )^4} \cdot\frac{x^{n+1}}{x^n}\right|$

. . $= \;\left|\,\frac{1}{6}\left(\frac{n}{n+1}\right)^4 x\,\right| \;=\; \left|\,\frac{1}{6}\left(\frac{1}{1 + \frac{1}{n}}\right)^4x\,\right|$

Then: . $\lim_{n\to\infty} \left|\,\frac{1}{6}\left(\frac{1}{1 + \frac{1}{n}}\right)^4x\,\right| \;=\;\left|\frac{1}{6}\cdot1^4\cdot x\,\right| \;=\;\left|\frac{x}{6}\right|$

So we have: . $\left|\frac{x}{6}\right| \:< \:1 \quad\Rightarrow\quad -6 \:<\:x\:<\:6$

. . The radius of convergence is: . $R \,=\,6$

At $x = \text{-}6\!:\;\;\sum^{\infty}_{n=1}\frac{(\text{-}6)^n}{6^nn^4} \;=\;\sum^{\infty}_{n=1}\frac{(\text{-}1)^n}{n^4}\quad\hdots$ an alternating $p$-series which converges

At $x = 6\!:\;\;\sum^{\infty}_{n=1}\frac{6^n}{6^nn^4} \;=\;\sum^{\infty}_{n=1}\frac{1}{n^4} \quad\hdots$ a convergent $p$-series

. . The interval of convergence is: . $[-6,\:6]$

• May 13th 2009, 05:58 AM
HallsofIvy
Quote:

Originally Posted by aaronb
I am finding the radius of convergence and interval of convergence.

the sum of n=1 to infinity of x^n/(6^n*n^4)

i used the ratio test on this and found that the limit goes to 1

I don't know what you mean by that. Do you mean that the limit of the ratios is less than 1 for |x|< 6?
Quote:

and the radius of convergence is 6.
Yes, that is correct.

Quote:

when x is -6, i know the series alternates but when i compare b of n to b of n+1 I don't see how b of n is greater than b of n+1 which is one of the stipulations for the alternating series test.
When x= -6, this is
$\sum_{n= 1}^\infty \frac{(-1)^}{6^n n^4}$
$\frac{\frac{1}{6^n n^4}{\frac{1}{6^{n+1}(n+1)^4}= 6\left(\frac{n+1}{n}\right)^4$
It should be clear that that is larger than 6 so the numerator is larger than the denominator- the sequence is decreasing.