Complex Power series: sum[i=0,infty]((2z-i)^4)/(n^4+1)

[xoninhas]

Hey everyone! How are you all doing? Hope you had a great summer! (for others winter )

I have here in my notes of Power Series (for complex numbers) two ways to calculate the radius of convergence and I'd like to know when to use each:

- \(\displaystyle \lim_{n->inf} \frac{a_n}{a_{n+1}}\)

- \(\displaystyle \frac{1}{ \lim_{n->inf} \sqrt[n]{|a_n|}}\) (the nth root I have no idea how to do in latex...

In the examples that use 1 or 2, here respectively:

\(\displaystyle \sum_{i=0}^{inf}{\frac{(2z-i)^4}{n^4 +1}}\) the radius of convergence is 1/2 and the region is |z - i/2| < 1/2 1st formula used!

The other uses the second and I don't know why:

\(\displaystyle \sum_{n=0}^{inf} \frac{(z +1 -i)^4}{n^4}\)

Thanks for any help!

 

[tkhunny]

Re: Complexes Power series

That's easy. Use the test that works and that you understand! <== Not entirely kidding.

\ s q r t [ n ] { stuff }

Produces

\(\displaystyle \sqrt[n]{stuff}\)

 

[Deleted member 4993]

Re: Complexes Power series

If the power of the numerator and the denominators are equal that is you can write those as:

\(\displaystyle (\frac{g(i)}{h(i)})^n\)

Then, generally, n_th root test is advisable.

 

[xoninhas]

Re: Complexes Power series

So if I understand correctly, both do the trick and should provide the same result... just for certain problems one is easier to use, and reach a result, than the other.

Can someone confirm this please?

Thanks again everyone!

 

[stapel]

Can someone confirm this please?

Try both, and see. :wink:

Eliz.

 

[Deleted member 4993]

Re: Complexes Power series

So if I understand correctly, both do the trick and should provide the same result... just for certain problems one is easier to use, and reach a result, than the other.

Yes - in general. More often than not, wrong choice will lead down the "inconclusive" garden path.

Can someone confirm this please?

Thanks again everyone!