Math Is Fun Forum

adam_dsutton

Member

Registered: 2007-09-12

Posts: 10

Find the Radius of convergence for the following power series....

I am having trouble finding the radius of convergence for:

I am not sure how to deal with the n!

All help is much appreciated!

P.s the sum is from n=0 to infinity...i didnt know how to do this using code

mikau

Member

Registered: 2005-08-22

Posts: 1,504

Re: Find the Radius of convergence for the following power series....

using the ratio test, divide the (n+1)th term by the n'th term, to get:

which reduces to:

now in the limit, as n approaches infinity (which is what we calculate when we apply the ratio test) the factor of (n)/(n+1) approaches 1, so all that we need be concerned with is x^(n! n), which is basically x^(infinity)

so the rule is, for what values of x will this limit have an absolute value less than 1? Intuitively, we can see that the answer is, -1 < x < 1. At these values, the high power will cause x to reduce itself, smaller and smaller and smaller until it eventually reaches zero.

What the ratio test doesn't tell you is what happens when the absolute value of the ratio is exactly 1, or in this case, where x = 1 or -1. (note the fact that both the ratio, and x must have an absolute value of 1 is purely coincidental)

so now we need to consider two specific case values: the series where x = 1, and the series where x = -1, that is:

and

my helpfulness stops here though, as I cannot remember how you work with these. But I'm pretty sure you can solve them using other methods. What you have to do is see if 1 and -1 are included in the radius of convergence, which will tell you if -1 and 1 can be included at the endpoints of the radius of convergence.

But actually, all this problem asks is the radius of convergence, which i believe does not require you to find out what happens at the very edge of the radius, only the overall size of the radius. So really, you don't need to worry about those two series.

Last edited by mikau (2008-05-31 14:58:23)

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A logarithm is just a misspelled algorithm.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Find the Radius of convergence for the following power series....

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

mikau

Member

Registered: 2005-08-22

Posts: 1,504

Re: Find the Radius of convergence for the following power series....

and how did you derive the second relation? thats kinda cool!

(edit) wait I get it! Every factorial after 2 has to be even! What a nifty trick!

Last edited by mikau (2008-06-01 07:34:23)

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A logarithm is just a misspelled algorithm.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Find the Radius of convergence for the following power series....

Right.  I'm surprised that you asked about the second and not the first.  Typically, the first is taken by definition, though it can be proven through use of any of the other equivalent definitions.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

mikau

Member

Registered: 2005-08-22

Posts: 1,504

Re: Find the Radius of convergence for the following power series....

well I recognized the first as being the maclauren expansion of e^x where x = 1.

I seem to remember there being an easy convergency test to show that 1/n! converges though. Just can't remember which.

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A logarithm is just a misspelled algorithm.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Find the Radius of convergence for the following power series....

Certainly 1/n! < 1/n^2, n >= 4

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

adam_dsutton

Member

Registered: 2007-09-12

Posts: 10

Re: Find the Radius of convergence for the following power series....

Cool, thanks for the help guys!