May 2nd, 2018, 12:50 PM  #11  
Senior Member Joined: Sep 2016 From: USA Posts: 520 Thanks: 293 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
\[ \limsup\limits_{n \to \infty} a_n^{\frac{1}{n}} = \limsup\limits_{n \to \infty} \left\left \left( \frac{\alpha n}{\alpha n +1} \right)^{n^2}i^n \right \right^{\frac{1}{n}} \] where $\left\left \cdot \right \right$ denotes the complex modulus. Secondly, you are not computing the limit \[ \lim\limits_{n \to \infty} \left\left \left( \frac{\alpha n}{\alpha n + 1} \right)^{n^2}i^n \right \right^{\frac{1}{n}} \] and in fact, you should not assume that it exists since this isn't true in general. This is the reason for working with the limsup (it always exists). All you need to apply the root test is to find the values of $\alpha$ which satisfy \[ \limsup\limits_{n \to \infty} \left\left \left( \frac{\alpha n}{\alpha n + 1} \right)^{n^2}i^n \right \right^{\frac{1}{n}} < 1. \] Here is a sketch which should help. 1. Note that for $a,b \in \mathbb{C}$, the modulus satisfies the property $ab = a\cdot b$ (prove this if it's something you haven't seen before). Extending this to exponentiation, you get $a^n = a^n$. From this you can conclude that \[ \limsup\limits_{n \to \infty} \left\left \left( \frac{\alpha n}{\alpha n + 1} \right)^{n^2}i^n \right \right^{\frac{1}{n}} = \limsup\limits_{n \to \infty} \left\left \frac{\alpha n}{\alpha n + 1}\right \right^{n}. \] 2. Suppose $ \limsup\limits_{n \to \infty} \left\left \frac{\alpha n}{\alpha n + 1}\right \right^{n} < 1$, then from the definition of limsup, there must be an $N \in \mathbb{N}$ such that for every $n > N$, we have $ \left\left \frac{\alpha n}{\alpha n + 1}\right \right^{n} < 1$ or equivalently, $ \left\left \frac{\alpha n}{\alpha n + 1}\right \right < 1$ (prove this). 3. Prove that if $z \in \mathbb{C}$, then $\frac{z}{z + 1} < 1$ if and only if $\text{Re}(z) > \frac{1}{2} $. Apply this to sequence in (2) to conclude that for large $n$, we have \[ \text{Re}(\alpha) > \frac{1}{2n} \implies \text{Re}(\alpha) > 0. \] 4. Now you can argue similarly, that if $\text{Re}(\alpha) \leq 0$, then \[ \limsup\limits_{n \to \infty} \left\left \frac{\alpha n}{\alpha n + 1} ^n \right \right \geq 1\] and conclude from the root test that the original series converges absolutely if and only if $\text{Re}(\alpha) > 0$. Last edited by skipjack; May 2nd, 2018 at 03:38 PM.  
May 2nd, 2018, 01:45 PM  #12 
Newbie Joined: May 2018 From: south africa Posts: 11 Thanks: 0 
I really do need to relook at this whole root test. Thank you for the help! Last edited by skipjack; May 2nd, 2018 at 03:32 PM. 

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