Radius of Convergence, limits: Complex numbers.

**ebb** · July 25th 2012, 07:33 AM

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 a_nzⁿ, where a_n = (2n + 1)!(n + 2i)ⁿ/(3n)!
My answers: 1. The limit does not exist. 2. R=27/4
Help would be veeeery much appreciated!

**Prove It** · July 25th 2012, 08:25 AM

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 a_nzⁿ, where a_n = (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 $\displaystyle \begin{align*} \sum{t_n} \end{align*}$ is convergent when $\displaystyle \begin{align*} \lim_{n \to \infty}\left|\frac{t_{n+1}}{t_n}\right| < 1 \end{align*}$ , so

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

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

So the radius of convergence is $\displaystyle \begin{align*} \frac{27}{4e} \end{align*}$ .

**ebb** · July 25th 2012, 10:35 PM

For 1. I'm not entirely sure what you mean by a 'counter of values'?
Thank you so much for your response!

**ebb** · July 25th 2012, 10:37 PM

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...

**Prove It** · July 26th 2012, 03:43 AM

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 $\displaystyle \begin{align*} \lim_{n \to \infty}\left|\left(\frac{n + 1 + 2i}{n + 2i}\right)^n\right| \end{align*}$ , so if that's the case you would probably have to use a CAS calculator.

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