Author Topic: TT1 Problem 2 (noon)  (Read 896 times)

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2466
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
TT1 Problem 2 (noon)
« on: October 19, 2018, 04:04:31 AM »
Determine the radius of convergence
(a) $\displaystyle{\sum_{n=0}^\infty \frac{3^n z^n }{n^2+1 }}$

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

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


oighea

  • Full Member
  • ***
  • Posts: 19
  • Karma: 21
  • I am a mulligan!
    • View Profile
Re: TT1 Problem 2 (noon)
« Reply #1 on: October 20, 2018, 06:07:32 PM »
(a) The radius is 1/3 (Ratio test).

$\displaystyle \lim_{n \rightarrow \infty} \left|\frac{3^{n+1}}{(n+1)^2 + 1} z^{n+1} / \frac{3^{n}}{n^2 + 1} z^{n}\right| = \lim_{n \rightarrow \infty}\left|\frac{3^{n+1}(n^2 + 1)}{3^n((n+1)^2+1)}\frac{z^{n+1}}{z^n}\right| = \lim_{n \rightarrow \infty}\left|\frac{3(n^2 + 1)}{n^2 + 2n + 2 + 1}z\right| = 3\lim_{n \rightarrow \infty}\left|\frac{n^2 + 1}{n^2 + 2n + 3}\right||z|$.

We now have a limit of a rational function, where both sides have the same degree and leading coefficients. Hence, we have $\displaystyle 3\lim_{n \rightarrow \infty}|1z|$.

The limit now evaluates to $|3z|$. We want that to be strictly less than 1 so the series converges.

$3|z| < 1 \implies |z| < \frac{1}{3}$. Hence, the radius of convergence is $\frac{1}{3}$. Shouldn't it be $|z|$? What happens as $|z|=\frac{1}{3}$? The inequality is not strict. It converges if $|z| = \frac{1}{3}$

(b) The radius is 0. The series only converges for z == 0.

$\displaystyle \lim_{n \rightarrow \infty} \left|\frac{((n + 1)!)^3}{(2(n+1))!}z^{n+1}/\frac{(n!)^3}{(2n)!}z^n\right| = \lim_{n \rightarrow \infty} \left|\frac{((n+1)!)^3(2n)!}{(n!)^3(2(n+1))!} \frac{z^{n+1}}{z^n}\right| = \lim_{n \rightarrow \infty} \left|\frac{((n+1)!)^3(2n)!}{(n!)^3(2n + 2)!}z\right| = \lim_{n \rightarrow \infty} \left|\frac{(n!(n+1))^3(2n)!}{(n!)^3(2n)!(2n+1)(2n+2)}z\right| = \lim_{n \rightarrow \infty} \left|\frac{(n!)^3(n+1)^3(2n)!}{(n!)^3(2n)!(2n+1)(2n+2)}\right||z|$

Cancelling out factorial terms on the fraction, we obtain $\displaystyle \lim_{n \rightarrow \infty} \left|\frac{(n+1)^3}{(2n+1)(2n+2)}\right||z|$. Note that the top of the rational function in n is cubic, and the bottom is quadratic. The limit at infinity of this rational function diverges, unless z is equal to 0.

Edit: Dealt with |z| on the boundary.
« Last Edit: October 21, 2018, 02:52:41 AM by oighea »

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2466
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: TT1 Problem 2 (noon)
« Reply #2 on: October 21, 2018, 01:44:28 AM »
As $|z-z_0|<R$ we claim that the series converges and as $|z-z_0|>R$ that it diverges. However, as $|z-z_0|=R$ neither root, nor ratio criteria provide an answer and one needs to use other criteria, which usually do not give us a uniform answer  (so, the series may converge for some $z\colon |z-z_0|=R$ and diverge for the rest of them. However in TT1 examples the answer is uniform and you need to find it

oighea

  • Full Member
  • ***
  • Posts: 19
  • Karma: 21
  • I am a mulligan!
    • View Profile
Re: TT1 Problem 2 (noon)
« Reply #3 on: October 21, 2018, 02:52:00 AM »
(a) The inequality $|z|$ being less than $\frac{1}{3}$ is not strict (Absolute Value, P-Series, Comparison)

Suppose $|z| = \frac{1}{3}$ - it is on the boundary of the radius, and $y = \mathrm{Im}\, z$.

Then we have $\displaystyle \sum_{n=0}^\infty \frac{3^nz^n}{n^2 + 1} = \sum_{n=0}^\infty \frac{3^n(|\frac{1}{3}|(\cos y + i \sin y))^n}{n^2 + 1} $

$\displaystyle= \sum_{n=0}^\infty \frac{3^n(\frac{1}{3})^n(\cos y + i \sin y)^n}{n^2 + 1}$

$\displaystyle= \sum_{n=0}^\infty \frac{1(\cos y + i \sin y)^n}{n^2 + 1} = \sum_{n=0}^\infty \frac{(\cos ny + i \sin ny)}{n^2 + 1}$ by De Moivré's Theorem.
Then, test absolute convergence:

$\displaystyle= \sum_{n=0}^\infty \frac{|(\cos ny + i \sin ny)|}{|n^2 + 1|}$

$\displaystyle= \sum_{n=0}^\infty \frac{1}{|n^2 + 1|} < \sum_{n=0}^\infty \frac{1}{|n^2|}$. Note $\displaystyle \lim_{n \to \infty}\left|\frac{1}{n^2 + 1}\right| = \lim_{n \to \infty}\left|\frac{1}{n^2}\right| = 0$

As the series is absolutely convergent when $|z| = \frac{1}{3}$, the inequality is not strict. Hence, the series converges for $|z| \leq \frac{1}{3}$ (Comparison and P-series).

(b) The series converges only when $|z| == 0$ - the inequality isn't strict. At that value, all the terms are zero.
« Last Edit: October 21, 2018, 02:04:00 PM by oighea »

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2466
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: TT1 Problem 2 (noon)
« Reply #4 on: October 21, 2018, 04:24:33 AM »
So, as $|z|=\frac{1}{3}$ (not $z=\frac{1}{3}$!) we got a series composed of modules $\sum_{n=0}^\infty \frac{1}{n^2+1}$. However $c_n\to 0$ is only necessary but not a sufficient criterium of the convergence. What criterium do you need to apply to prove convergence?

oighea

  • Full Member
  • ***
  • Posts: 19
  • Karma: 21
  • I am a mulligan!
    • View Profile
Re: TT1 Problem 2 (noon)
« Reply #5 on: October 21, 2018, 02:07:22 PM »
Absolute convergence: I proved $\displaystyle= \sum_{n=0}^\infty \frac{1}{|n^2 + 1|} < \sum_{n=0}^\infty \frac{1}{|n^2|}$ is absolutely convergent for $|z| = \frac{1}{3}$. It is also a p-series, and uses a comparison, and since the power is greater than 1, it is absolutely convergent.