Radius of Convergence for a Power Series with Ratio Test

[Saladsamurai]

Homework Statement

Determine the radius of Convergence using the ratio test of:

\sum_o^{\infty}\frac{n^6}{3^n+n}(x+4)^{8n+1}\qquad(1)

Homework Equations

R = \frac{1}{\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|}\qquad(2)

The Attempt at a Solution

Ok. In order to use (2), we must first put (1) into standard form: \sum a_n(x - x_o)^n.

I am following a hint that I should let m = 8n +1 however I am not sure what to do with the summation limits? If m = 8n +1, then at n = 0, m = 1. So do I just run the sum from 1 to \infty? And replace n everywhere with n = (m - 1)/8 ?

Thanks!

 

[jackmell]

I don't understand why we can't just consider the convergence of the series:

\sum_{n=0}^{\infty} a_n

with:

a_n=\frac{n^6}{3^n+n}(x+4)^{8n+1}

and using the radio test with:

\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|

If that's valid, I get:

R=3^{1/8}

 

[Saladsamurai]

Homework Statement

Determine the radius of Convergence using the ratio test of:

\sum_o^{\infty}\frac{n^6}{3^n+n}(x+4)^{8n+1}\qquad(1)

Homework Equations

R = \frac{1}{\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|}\qquad(2)

The Attempt at a Solution

Ok. In order to use (2), we must first put (1) into standard form: \sum a_n(x - x_o)^n.

I am following a hint that I should let m = 8n +1 however I am not sure what to do with the summation limits? If m = 8n +1, then at n = 0, m = 1. So do I just run the sum from 1 to \infty? And replace n everywhere with n = (m - 1)/8 ?

Thanks!

This is geting silly:

I have that

<br /> \frac{a_{m+1}}{a_m} = \left(\frac{m}{8}\right)^6\left(\frac{8}{m-1}\right)^6<br /> \left(\frac{3^{\frac{m-1}{8}}+\frac{m-1}{8}}{3^{m/8}+m/8}\right)<br />

= \frac{a_{m+1}}{a_m} = \left(\frac{m}{m-1}\right)^6\left(\frac{3^{\frac{m-1}{8}}+\frac{m-1}{8}}{3^{m/8}+m/8}\right)<br />

I cannot see how to reduce this further. Any ideas? I'm lookin' at you jackmell

 

[jackmell]

I don't understand why we can't just consider the convergence of the series:

\sum_{n=0}^{\infty} a_n

with:

a_n=\frac{n^6}{3^n+n}(x+4)^{8n+1}

and using the radio test with:

\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|

If that's valid, I get:

R=3^{1/8}

I'm not entirely sure about that. Also, numerically, looks like the radius of convergence appears to be one over that:

\frac{1}{3^{1/8}}

but I can't seem to get to that value.

 

[jackmell]

= \frac{a_{m+1}}{a_m} = \left(\frac{m}{m-1}\right)^6\left(\frac{3^{\frac{m-1}{8}}+\frac{m-1}{8}}{3^{m/8}+m/8}\right)<br />

I cannot see how to reduce this further. Any ideas? I'm lookin' at you jackmell

We can do that. That first term has a limit of one right (it approaches m/m)^6). Now look at the (m-1)/8 in the numerator and the m/8 in the denominator. Both of those are O(m) but pale in comparison to the other terms having exponential order so for very large m, we can neglect them and just consider:

\lim_{m\to\infty} \frac{3^{1/8(m-1)}}{3^{m/8}} but when we add the exponents we get:

\frac{m-1}{8}-\frac{m}{8}=-1/8 so that:

\lim_{m\to\infty} \frac{3^{1/8(m-1)}}{3^{m/8}}=3^{-1/8}

 

[Saladsamurai]

We can do that. That first term has a limit of one right (it approaches m/m)^6). Now look at the (m-1)/8 in the numerator and the m/8 in the denominator. Both of those are asymptotic to m. But both pale in comparison to the other terms so for very large m, we can neglect them and just consider:

\lim_{m\to\infty} \frac{3^{1/8(m-1)}}{3^{m/8}} but when we add the exponents we get:

\frac{m-1}{8}-\frac{m}{8}=-1/8 so that:

\lim_{m\to\infty} \frac{3^{1/8(m-1)}}{3^{m/8}}=3^{-1/8}

Ok great. I follow your qualitative description just fine. Now I woud like to flesh out the details using appropriate limit rules.

We have a limit of a product which equals the product of the limits. The first factor

\left(\frac{m}{m-1}\right)^6

is a rational function of two polynomials of the same order, so its limit is simply the ratio of the coefficients of the leading terms in numerator and denominator: 1/1 = 1.

Now as for the second factor:

\left(\frac{3^{\frac{m-1}{8}}+\frac{m-1}{8}}{3^{m/8}+m/8}\right)

I am trying to pinpoint a nice rule that I can apply to back the statement that the exponentials are 'more important.'

Any ideas? Thanks for your help so far

 

[jackmell]

It's all a matter of order, or big-O:

O(3^n)+O(n)\sim O(3^n)

That is, for very big n, that sum asymptotically approaches the 3^{n}[/tex] term as the smaller-ordered term becomes less and less significant.<br /> <br /> So that:<br /> <br /> \lim_{n\to\infty}\left(\frac{O(3^n)+O(n)}{O(3^{n-1})+O(n)}\right)\sim \frac{O(3^{n})}{O(3^{n-1})}<br /> <br /> Essentially, as n grows very large, only the highest-ordered terms matter.<br /> <br /> I may not have that exactly right but I think it&#039;s close.