Author Topic: TT1 Problem 2 (morning)  (Read 604 times)

Victor Ivrii

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TT1 Problem 2 (morning)
« on: October 19, 2018, 03:53:25 AM »
Determine the radius of convergence

(a) $\displaystyle{\sum_{n=1}^\infty \frac{2^n z^n n}{3^n }}$

(b) $\displaystyle{\sum_{n=1}^\infty \frac{z^n n!}{ (2n)! }}$

If the radius of convergence is $R$, $0<R< \infty$, determine for each  $z\colon |z|=R$ if this series converges.

Vedant Shah

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Re: TT1 Problem 2 (morning)
« Reply #1 on: October 19, 2018, 08:59:45 AM »
(a)
By Ratio Test:
$\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| = \lim_{n \to \infty} |\frac{2^{n+1} z^{n+1} {n+1}}{3^{n+1}} \frac{3^{n}}{2^{n} z^{n} n}| \\
=\lim_{n \to \infty}  |\frac{n+1}{n} \frac{2}{3} z| \\
= |\frac{2}{3} z|
$

We want $\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| < 1$:
$
|\frac{2}{3} z| < 1 \\
|z| < \frac{3}{2} \\
R = \frac{3}{2}
$

Now testing $|z| = \frac{3}{2}$:
$
\sum\limits_{n=1}^{\infty} |\frac{2^{n} z^{n} {n}}{3^{n}}| \\
= \sum\limits_{n=1}^{\infty} \frac{|2|^{n} |z|^{n} {|n|}}{|3|^{n}} \\
= \sum\limits_{n=1}^{\infty} \frac{|2|^{n} |\frac{3}{2}|^{n} {|n|}}{|3|^{n}} \\
= \sum\limits_{n=1}^{\infty} |n|
$

The series diverges at the boundary, thus we have:
$|z| < \frac{3}{2}$

(b)
By Ratio Test:
$
\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| = \lim_{n \to \infty} |\frac{ {(n+1)}! z^{n+1} }{{(2n + 2)}!} \frac{{(2n)}!}{{n}! z^{n}}| \\
=\lim_{n \to \infty}  |\frac{n+1}{(2n+1)(2n+2)}z| \\
= 0
$

Thus, $\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| < 1 \forall z$
Therefore, we have the series converges for:
$
R = \infty \\
|z| < \infty
$