Toronto Math Forum
MAT3342018F => MAT334Tests => Term Test 1 => Topic started by: Victor Ivrii 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.

(a) The radius is 1/3 (Ratio test).
$\displaystyle \lim_{n \rightarrow \infty} \left\frac{3^{n+1}}{(n+1)^2 + 1} z^{n+1} / \frac{3^{n}}{n^2 + 1} z^{n}\right = \lim_{n \rightarrow \infty}\left\frac{3^{n+1}(n^2 + 1)}{3^n((n+1)^2+1)}\frac{z^{n+1}}{z^n}\right = \lim_{n \rightarrow \infty}\left\frac{3(n^2 + 1)}{n^2 + 2n + 2 + 1}z\right = 3\lim_{n \rightarrow \infty}\left\frac{n^2 + 1}{n^2 + 2n + 3}\rightz$.
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.
$3z < 1 \implies z < \frac{1}{3}$. 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.
$\displaystyle \lim_{n \rightarrow \infty} \left\frac{((n + 1)!)^3}{(2(n+1))!}z^{n+1}/\frac{(n!)^3}{(2n)!}z^n\right = \lim_{n \rightarrow \infty} \left\frac{((n+1)!)^3(2n)!}{(n!)^3(2(n+1))!} \frac{z^{n+1}}{z^n}\right = \lim_{n \rightarrow \infty} \left\frac{((n+1)!)^3(2n)!}{(n!)^3(2n + 2)!}z\right = \lim_{n \rightarrow \infty} \left\frac{(n!(n+1))^3(2n)!}{(n!)^3(2n)!(2n+1)(2n+2)}z\right = \lim_{n \rightarrow \infty} \left\frac{(n!)^3(n+1)^3(2n)!}{(n!)^3(2n)!(2n+1)(2n+2)}\rightz$
Cancelling out factorial terms on the fraction, we obtain $\displaystyle \lim_{n \rightarrow \infty} \left\frac{(n+1)^3}{(2n+1)(2n+2)}\rightz$. 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.

As $zz_0<R$ we claim that the series converges and as $zz_0>R$ that it diverges. However, as $zz_0=R$ 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 $z\colon zz_0=R$ and diverge for the rest of them. However in TT1 examples the answer is uniform and you need to find it

(a) The inequality $z$ being less than $\frac{1}{3}$ is not strict (Absolute Value, PSeries, 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 MoivrÃ©'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 Pseries).
(b) The series converges only when $z == 0$  the inequality isn't strict. At that value, all the terms are zero.

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?

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 pseries, and uses a comparison, and since the power is greater than 1, it is absolutely convergent.