# Radius of Convergence, limits: Complex numbers.

• Jul 25th 2012, 07:33 AM
ebb
Radius of Convergence, limits: Complex numbers.
I'm really shaky with radius of convergence and limits when complex numbers are thrown into the mix, so I would be really grateful if someone could check I've answered these questions correctly! (I can't find any examples online that are quite as complicated)
1. Find the limit of (2n*e(( ln(n^2) + i*pi*n )/(( 16(n^2) + 5i ))^0.5))/((4n2 + 3in)(1/2)) [From n=1 to infinity]
2. Compute the radius of convergence of the power series: (sum from n=1 to infinity) of anzn, where an = (2n + 1)!(n + 2i)n/(3n)!
My answers: 1. The limit does not exist. 2. R=27/4
Help would be veeeery much appreciated!
• Jul 25th 2012, 08:25 AM
Prove It
Re: Radius of Convergence, limits: Complex numbers.
Quote:

Originally Posted by ebb
I'm really shaky with radius of convergence and limits when complex numbers are thrown into the mix, so I would be really grateful if someone could check I've answered these questions correctly! (I can't find any examples online that are quite as complicated)
1. Find the limit of (2n*e(( ln(n^2) + i*pi*n )/(( 16(n^2) + 5i ))^0.5))/((4n2 + 3in)(1/2)) [From n=1 to infinity]
2. Compute the radius of convergence of the power series: (sum from n=1 to infinity) of anzn, where an = (2n + 1)!(n + 2i)n/(3n)!
My answers: 1. The limit does not exist. 2. R=27/4
Help would be veeeery much appreciated!

For question 1, you need to find a limit to a particular value, not to a counter of values, which you have written.

For question 2, a series \displaystyle \displaystyle \begin{align*} \sum{t_n} \end{align*} is convergent when \displaystyle \displaystyle \begin{align*} \lim_{n \to \infty}\left|\frac{t_{n+1}}{t_n}\right| < 1 \end{align*}, so

\displaystyle \displaystyle \begin{align*} \lim_{n \to \infty}\left| \frac{ \frac{[2(n+1) + 1]!(n + 1 + 2i)^{n + 1}}{[3(n+1)]!} \, z^{n+1} }{ \frac{(2n+1)!(n + 2i)^n}{(3n)!}\,z^n } \right| &< 1 \\ \lim_{n \to \infty}\left| \frac{\frac{(2n + 3)!(n + 1 + 2i)(n + 1 + 2i)^n}{(3n + 3)!}}{\frac{(2n + 1)!(n + 2i)^n}{(3n)!}} \right||z| &< 1 \\ \lim_{n \to \infty}\left| \frac{(3n)!(2n+3)!(n + 1 + 2i)(n + 1 + 2i)^n}{(3n + 3)!(2n + 1)!(n + 2i)^n} \right||z| &< 1 \\ \lim_{n \to \infty}\left| \frac{(2n + 3)(2n + 2)(n + 1 + 2i)(n + 1 + 2i)^n}{(3n + 3)(3n + 2)(3n + 1)(n + 2i)^n} \right||z| &< 1 \\ \lim_{n \to \infty}\left| \frac{(2n + 3)(2n + 2)(n + 1 + 2i)}{(3n + 3)(3n + 2)(3n + 1)} \right| \lim_{n \to \infty}\left| \left(\frac{n + 1 + 2i}{n + 2i}\right)^n \right| |z| &< 1 \\ \lim_{n \to \infty} \left| \frac{4n^3 + (14 + 8i)n^2 + (16 + 20i)n + 6 + 12i}{27n^3 + 54n^2 + 33n + 6} \right| \lim_{n \to \infty}\left| e^{n\ln{\left(\frac{n + 1 + 2i}{n + 2i}\right)}} \right| |z| &< 1 \end{align*}

\displaystyle \displaystyle \begin{align*} \lim_{n \to \infty}\left| \frac{4 + \frac{14 + 8i}{n} + \frac{16 + 20i}{n^2} + \frac{6 + 12i}{n^3}}{27 + \frac{54}{n} + \frac{33}{n^2} + \frac{6}{n^3}} \right| \lim_{n \to \infty}\left|e^{\frac{\ln{\left( 1 + \frac{1}{n + 2i}\right)}}{\frac{1}{n}}}\right||z| &< 1 \\ \frac{4}{27}\lim_{n \to \infty}\left|e^{\frac{\frac{1}{n + 1 + 2i} - \frac{1}{n + 2i}}{-\frac{1}{n^2}}}\right||z| &< 1 \textrm{ by L'Hospital's Rule} \\ \frac{4}{27}\lim_{n \to \infty}\left| e^{\frac{3 - 4i}{n + 1 + 2i} - \frac{4}{n + 2i} + 1} \right||z| &< 1 \\ \frac{4e}{27} |z| &< 1 \\ |z| &< \frac{27}{4e} \end{align*}

So the radius of convergence is \displaystyle \displaystyle \begin{align*} \frac{27}{4e} \end{align*}.
• Jul 25th 2012, 10:35 PM
ebb
Re: Radius of Convergence, limits: Complex numbers.
For 1. I'm not entirely sure what you mean by a 'counter of values'?
Thank you so much for your response!
• Jul 25th 2012, 10:37 PM
ebb
Re: Radius of Convergence, limits: Complex numbers.
Also, in the module I'm doing we've not been taught L'Hopital's rule, although I've come across it in further reading, and the lecturer said we will not be allowed to answer questions using it in the exam...
• Jul 26th 2012, 03:43 AM
Prove It
Re: Radius of Convergence, limits: Complex numbers.
Quote:

Originally Posted by ebb
Also, in the module I'm doing we've not been taught L'Hopital's rule, although I've come across it in further reading, and the lecturer said we will not be allowed to answer questions using it in the exam...

Well I can't think of another way to evaluate \displaystyle \displaystyle \begin{align*} \lim_{n \to \infty}\left|\left(\frac{n + 1 + 2i}{n + 2i}\right)^n\right| \end{align*}, so if that's the case you would probably have to use a CAS calculator.