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May 2nd, 2018, 12:50 PM   #11
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Originally Posted by niemi View Post
I need help with the following, but I am stuck. Is the i to the power n a typo? Please help me solve this question...

"Use the nth-root test to find the conditions on the constant a in order that the series

$\displaystyle \sum _{n=1}^{\infty \:}\left(\frac{an}{an+1}\right)^{n^2}i^n$

is absolutely convergent
You should reread the root test. You are misunderstanding quite a few things. First off, as mentioned you need to take the modulus of the sequence. This gives you a sequence of reals to works with. Namely, you are interested in the sequence
\[ \limsup\limits_{n \to \infty} ||a_n||^{\frac{1}{n}} = \limsup\limits_{n \to \infty} \left|\left| \left( \frac{\alpha n}{\alpha n +1} \right)^{n^2}i^n \right| \right|^{\frac{1}{n}} \]
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 $\text{Re}(\alpha) > 0$.
Thanks from niemi

Last edited by skipjack; May 2nd, 2018 at 03:38 PM.
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May 2nd, 2018, 01:45 PM   #12
Joined: May 2018
From: south africa

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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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