MAT334-2018F > Term Test 1

TT1 Problem 2 (morning)

(1/1)

Victor Ivrii:

(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:
(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:
$R = \infty \\ |z| < \infty$