For every \(n \in \mathbb N\), let \(f_n(x)=\frac{1}{x^n}\) for \(x\in[1,\infty)\). Show that \(\{f_n\}_{n\in \mathbb N}\) converges pointwise on \([1,\infty)\), and determine whether convergence is uniform on \([1,\infty)\). For a fixed \(R>1\), determine whether convergence is uniform on \([R,\infty)\).
Let \[f(x)=\begin{cases} 1&\text{ if }x=1\\ 0&\text{ if }x>1 \end{cases}\] Clearly \(f_n\rightarrow f\) pointwise on \([1,\infty)\), because \(f_n(1)=1\) for all \(n\), and if \(x>1\) then \(\frac{1}{x^n}\rightarrow 0\).
[In full epsilonic detail, for every \(\epsilon \in (0,1)\) and \(x>1\) we may pick any \(N>|\ln(\epsilon)| /\ln(x)\), because then, for \(n > N\), we have \(|f_n(x)-f(x)| = 1/x^n < 1/x^N < \epsilon .\:\)]
Method 1 (using a big theorem):Each \(f_n\) is continuous on \([1,\infty)\) and \(f\) is not, so by the Uniform Limit Theorem the convergence cannot be uniform on \([1,\infty)\).
Method 2 (by lesser means): Pick any real
number \(c >0\) and consider the
sequence \(x_n\) = \((1+c)^{1/n}\) in \([1, \infty)\). We have \(|f_n(x_n)-f(x_n)| = |1/(1+c) - 0| =
1/(1+c)\), so by the test for non-uniform convergence convergence
is not uniform.
For \([R,\infty)\), note that for any
\(x\in[R,\infty)\) we have \[|f_n(x)-f(x)|=\frac{1}{x^n}\leq
\frac{1}{R^n}\rightarrow0 \text{ as }n\rightarrow\infty.\] So,
according to a lemma from class the convergence is uniform in \([R,\infty)\).
[If you wish you can prove all of the above directly from the definitions without appealing to any of the Theorems/Lemmas.]
For every \(n \in \mathbb N\), let \(f_n(x)=\arctan(nx)\) for \(x\in \mathbb R\). Show that \(\{f_n\}_{n\in\mathbb N}\) converges pointwise on \(\mathbb R\) to \[f(x) = \begin{cases} \pi/2, & \quad \text{if } x > 0.\\ 0, & \quad \text{if } x =0.\\ -\pi/2, &\quad \text{if } x < 0.\ \end{cases}\] Is the convergence uniform?
In full epsilonic detail: Let \(\epsilon>0\) and pick \(x \in \mathbb R\). If \(x=0\) we have \(f_n(x) = \arctan(0) = 0\), and so convergence to \(0\) is trivial. If \(x>1\), then pick any \(N > \tan(\pi/2-\epsilon)/x\). Then for \(n > N\) we have \[|f_n(x) - f(x)| \:= \:\pi/2- \arctan(nx) \: < \:\pi/2- \arctan(x\tan(\pi/2-\epsilon)/x) \:=\: \pi/2 - (\pi/2-\epsilon)\: =\: \epsilon,\] since \(\arctan\) is an increasing function. Similarly, if \(x<0\) pick any \(N > \tan(\epsilon-\pi/2)/x\) (which is positive). Then for \(n > N\) we have \[|f_n(x) - f(x)| \:= \:\arctan(nx) + \pi/2 \: < \:\arctan(x\tan(\epsilon - \pi/2)/x) + \pi/2 \:=\: (\epsilon-\pi/2) + \pi/2 \: =\: \epsilon.\] Note that each \(N\) depends on \(x\) so we do not expect uniform convergence. Indeed:
Method 1 (using a big theorem):Each \(f_n\) is continuous on \(\mathbb R\) and \(f\) is not, so by the Uniform Limit Theorem convergence cannot be uniform on \(\mathbb R\).
Method 2 (by lesser means): Pick any real number \(0 < c< \pi/2\) and consider the sequence \(x_n\) = \(\tan(c)/n\) in \(\mathbb R\). We have \(|f_n(x_n)-f(x_n)| = |c - \pi/2| = \pi/2-c\), so by the test for non-uniform convergence the convergence is not uniform.
Show that for any \(\rho>0\) the sequence \(\bigl\{\frac{1}{nz}\bigr\}_{n\in\mathbb N}\) converges uniformly on \(\{\,z \in \mathbb C: \ {\left|z\right|}\geq \rho\,\}\).
Does \(\bigl\{\frac{1}{nz}\bigr\}_{n\in\mathbb N}\) converge uniformly on \(\mathbb C^\ast:=\mathbb C\setminus \{ 0\}\)?
For every \(z\) in the set we have \(|1/(nz)| = |z|^{-1}(1/n) \to 0\) as \(n \to \infty\), so the pointwise limit is the constant function \(f(z) = 0\).
For each fixed \(n\) we see that \(\bigl|\frac{1}{nz}\bigr| \le \frac{1}{\rho n}\) for all \(|z|\ge \rho\). Since \(\bigl\{\frac{1}{\rho n}\bigr\}\to 0\) as \(n\to\infty\), it follows from a lemma from class that the given sequence converges uniformly to the function \(f(z)=0\) on the set \(\{\,z \in \mathbb C: \ {\left|z\right|}\geq \rho\,\}\).
The pointwise limit on \(\mathbb C^\ast\) is the constant function \(f(z)=0\), so, if convergence is uniform then the uniform limit must be \(f(z)=0\). However, for any \(c>0\) consider the sequence \(z_n = c/n\) in \(\mathbb C^\ast\). We have \(|f_n(z_n) - f(z_n) | = 1/c\), so by the test for non-(uniform convergence) the convergence is not uniform.
[Alternatively, by hand: for any fixed \(n\), as \({z\to 0}\) we see that \(\bigl|\frac{1}{nz}\bigr|\) tends to infinity. So given \(\epsilon >0\) and any \(n \in \mathbb N\), we can always find a point \(z \in \mathbb C\) such that \(\bigl|\frac{1}{nz}\bigr| \geq \epsilon\); namely \(z=1/(n\epsilon)\) will do. Hence convergence is not uniform.]
For any \(\rho > 0\), show that \(\bigl\{\frac{n}{1+nz}\bigr\}_{n\in\mathbb N}\) converges uniformly on \(\{\,z \in \mathbb C: \ {\left|z\right|} > \rho \,\}\). Does it converge uniformly on \(\mathbb C^{\ast}\,\)?
Given any fixed \(z \in \{z \in \mathbb C:
\ {\left|z\right|} > \rho \,\}\) we have \(\lim_{n\to\infty} \frac{n}{1+nz} =
\lim_{n\to\infty} \frac{1}{(1/n)+z} = \frac 1z\), so the
pointwise limit on the set is the function \(f(z)=1/z\). Also, for each fixed \(n\), we see that \[\left|
\frac{n}{1+nz}-\frac{1}{z}\right|=\left|\frac{1}{z(1+nz)}\right|.\]
However, for \(|z| > \rho\) and
\(n\) sufficiently large (\(n > 1/\rho\), say), we have that \(|z(1+nz)| > \rho|1+nz|\ge \rho (\rho
n-1)\), so that \[\left|f_n(z)-f(z)\right| =
\left|\frac{n}{1+nz}-\frac{1}{z}\right| \le \frac{1}{\rho(\rho
n-1)}.\] Let \(\epsilon>0\).
Since \(\bigl\{\frac{1}{\rho(\rho
n-1)}\bigr\}\to 0\) as \(n\to
\infty\), we can find \(N \in\mathbb
N\) such that \(\frac{1}{\rho(\rho
n-1)} < \epsilon\) for all \(n>N\); this proves that converge is
uniform (to the function \(f(z)=1/z\))
on \(\{z \in \mathbb C: \ {\left|z\right|}
> \rho \,\}\).
To see if the convergence is not uniform on all of \(\mathbb C^{\ast}\) we wish to construct a
sequence \(z_n\) in \(\mathbb C^\ast\) and find a constant \(c>0\) such that \(|f_n(z_n)-f(z_n)|=c \,\). Let’s just find
such a sequence of strictly positive real numbers. Let \(c>0\). Then we want \(z_n\) such that \(\frac{1}{z_n(1+nz_n)}=c\). By rearanging
this equation and completing the square we see that \[z_n = \frac{1}{2n} + \sqrt{\frac{1}{4n^2}+
\frac{1}{cn}}\] does the trick. Each of these is clearly in \(\mathbb C^{\ast}\), so convergence is not
uniform.
[Alternatively: To show convergence is not uniform on all of \(\mathbb C^{\ast}\) we can argue more directly. Given \(n\), we can always find a point \(z \in \mathbb C^{\ast}\) such that \(1+nz =0\) (namely, \(z=-1/n\)). Hence, nearby \(z=-1/n\), the difference \(\bigl|\frac{1}{z(1+nz)}\bigr|\) becomes arbitrarily large. That is, for any \(\epsilon>0\) and any \(n \in \mathbb N\) we can find \(z\) such that \(\left|f_n(z)-f(z)\right| = \bigl|\frac{1}{z(1+nz)}\bigr| \geq \epsilon\).]
Show that if \(0<\rho<1\), then the sequence \(\bigl\{\frac{1}{1+z^n}\bigr\}_{n\in\mathbb N}\) converges uniformly on \(\{\,z \in \mathbb C: \ {\left|z\right|}\leq \rho\,\}\) to the constant function \(f(z)=1\). On the other hand, show that the sequence converges uniformly on \(\{\,z \in \mathbb C: \ {\left|z\right|}\geq \rho^{-1}\,\}\) to the constant function \(f(z)=0\).
Show that the sequence \(\bigl\{\frac{1}{1+z^n}\bigr\}_{n\in\mathbb N}\) does not converge uniformly on \(\mathbb{D} = \{\,z \in \mathbb C: \ {\left|z\right|} < 1\,\}\).
(i) Let \(f_n(z)=1/(1+z^n)\) be defined on \(\{z \in \mathbb C:{\left|z\right|}\le\rho\}\) and let \(f(z)=1\). Then, since by the reverse triangle inequality \(|1+z^n|=|1- (-z^n)| \geq \bigl|\: |1|- |(-z^n)|\:\bigr| = 1-|z|^n\) for \(z\) with \(|z| < 1\), we have for \(z\) in \(\{z \in \mathbb C:{\left|z\right|}\le\rho\}\) that \[{\left|f_n(z)-f(z)\right|}\quad = \quad\left|\frac{1}{1+z^n}-1\right| \quad=\quad \left|\frac{z^n}{1+z^n}\right| \quad \le \quad\frac{{\left|z\right|}^n}{1-{\left|z\right|}^n} \quad \le\quad\frac{\rho^n}{1-\rho^n} \to 0 \hbox{ \ as } n\to \infty .\] Hence \(\{f_n\}\) converges uniformly on \(\{z \in \mathbb C:{\left|z\right|}\le\rho\}\) to the constant function \(f(z)=1\).
Now consider points in \(\{\,z \in \mathbb
C: \ {\left|z\right|}\geq \rho^{-1}\,\}\) . Here \(|z|>1\), so \(|1+z^n|=|1- (-z^n)| \geq \bigl||\: |1|-
|(-z^n)|\:\bigr| = |z|^n-1\). Then \[{\left|f_n(z)-f(z)\right|}\quad=\quad
\left|\frac{1}{1+z^n}\right| \quad\le \quad
\frac{1}{{\left|z\right|}^n-1}\quad\le\quad
\frac{1}{\rho^{-n}-1}\quad=\quad\frac{\rho^n}{1-\rho^n} \to 0 \hbox{ \
as } n\to \infty .\] Therefore \(\{f_n\}\) converges uniformly on \(\{\,z \in \mathbb C: \ {\left|z\right|}\geq
\rho^{-1}\,\}\) to the constant function \(f(z)=0\).
(ii) As above, the sequence converges pointwise on \(\{z:{\left|z\right|}<1\}\) to the
constant function \(f(z)=1\). To show
convergence is not uniform it is enough to find \(c>0\) and a sequence \(z_n\) in \(\mathbb D\) such that \(|f_n(z_n)-f(z_n)| = c\). Let’s just find
such a sequence \(z_n\) in the open
unit interval \((0,1)\) on the real
line. If \[|f_n(z_n)-f(z_n)| =
\frac{|(z_n)^n|}{|1+(z_n)^n|} = \frac{(z_n)^n}{1+(z_n)^n} = c,\]
then rearranging gives us \(z_n =
\left(\frac{c}{1-c}\right)^{1/n}\). So, we may for example choose
\(c=1/3\), then \(z_n = \left(\frac{c}{1-c}\right)^{1/n} = \frac {1}
{2^{1/n}} \in \mathbb D\) and \(|f_n(z_n)-f(z_n)| = 1/3\), so the
convergence is not uniform in \(\mathbb
D\).
[Alternatively: Note that for a fixed \(n\), we have \[\lim_{z\to 1} |f_n(z)-f(z)|=\lim_{z\to 1}\frac{|z^n|}{|1+z^n|}=\frac{1}{1+1}=\frac{1}{2}\ .\] So let \(\epsilon= 1/4\), say. Then for any \(n\) we can find \(z \in \mathbb D\) such that \(|f_n(z)-f(z)| \geq \frac 1 2 - \epsilon = \epsilon\). Hence the convergence is not uniform.]
For every \(n \in \mathbb N\), let \(f_n(z)=\sin(z/n)\) for \(z\in \mathbb C\). Show that \(\{f_n\}_{n\in\mathbb N}\) converges pointwise on \(\mathbb C\). Let \(\rho\) be a positive real number. Show that \(\{f_n\}_{n\in\mathbb N}\) converges uniformly on \(\{z:{\left|z\right|}\le \rho\}\). Show that \(\{f_n\}_{n\in\mathbb N}\) does not converge uniformly on \(\mathbb C\).
For fixed \(z\), \(\lim_{n\to\infty}z/n=0\), so, since \(\sin(z)\) is continuous at \(z=0\), it follows that, for fixed \(z\), \(\lim_{n\to\infty}\sin(z/n)=\sin(0)=0\). Hence \(\{f_n\}\) converges pointwise on \(\mathbb C\) to the zero function \(f(z)=0\).
For each fixed \(n\), \[|\sin(z/n) - f(z)|=|\sin(x/n)\cosh(y/n)+i\cos(x/n)\sinh(y/n)|\le |\sin(x/n)\cosh(y/n)|+|\sinh(y/n)|,\] so, for \(|z| \le \rho\), \[|\sin(z/n) - f(z)|\le (\rho/n) \cosh (\rho/n)+\sinh(\rho/n).\] Putting \(s_n\) equal to the RHS of the above inequality, we see that \(\lim_{n\to\infty}s_n=0\). Hence, by a lemma from class, we have that \(\{f_n\}\) converges uniformly to the zero function on \(\{z:{\left|z\right|}\le\rho\}\).
[Alternatively (without using the Lemma): Note \(|\sin(z/n) - f(z)|\) can be made arbitraily small for large \(n\) independent of \(z\), that is, for any \(\epsilon>0\) we can find \(N \in \mathbb N\) such that for \(n>N\) we have \(s_n < \epsilon\), so \(\{f_n\}\) converges uniformly to the zero function on \(\{z:{\left|z\right|}\le\rho\}\).]
To see if the convergence is not uniform on \(\mathbb C\) we wish to construct a sequence \(\left\lbrace z_n\right\rbrace_{n\in\mathbb N}\) and find a constant \(c>0\) such that \(|f_n(z_n)-f(z_n)|=c \,\). The sequence \(z_n=in\) with \(c=\sinh(1)>0\) does the trick, for then \(|f_n(z_n)-f(z_n)| = |\sin(z_n/n)|= |\sin(i)|=\sinh(1)\). Thus, convergence is not uniform.
[Alternatively, just say that for fixed \(n\), we have \(lim_{y\to \infty} |\sin(iy/n)|= lim_{y\to \infty} |\sinh(y/n)|=\infty\). Hence for every \(n\), \(|f_n(z)- f(z)|\) is unbounded. So convergence isn’t uniform on \(\mathbb C\).]
For every \(n \in \mathbb N\), let \(f_n(x)=\cos\left(1+\frac{x}{n}\right)\) for \(x\in\mathbb{R}\). Show that \(\{f_n\}_{n\in\mathbb N}\) converges pointwise and determine whether convergence is uniform on \(\mathbb{R}\). For fixed \(R>0\), is the convergence uniform on \([0,R]\)?
For fixed \(x\), \(\lim_{n\to\infty}x/n=0\), so, since \(\cos(x)\) is continuous at \(x=1\), it follows that, for fixed \(x\) we have \(\lim_{n\to\infty}\cos(1+x/n)=\cos(1)\). Hence \(\{f_n\}\) converges pointwise on \(\mathbb R\) to the constant function \(f(x)=\cos(1)\).
To see if the convergence is not uniform on \(\mathbb R\) we wish to construct a sequence \(x_n\) in \(\mathbb R\) and find a constant \(c>0\) such that \(|f_n(x_n)-f(x_n)|=c \,\). The sequence \(x_n=(\frac{\pi}{2}-1)n\) with \(c=\cos(1)>0\) does the trick, for then \(|f_n(x_n)-f(x_n)| = |\cos(1+x_n/n)-\cos(1)|=| \cos(\pi/2) - \cos(1) | = \cos(1)\). Thus, convergence is not uniform.
[Alternatively: Note that for fixed \(n\), \(\cos\left(1+\frac{x}{n}\right)\) periodically takes value \(0\) as \(x\rightarrow\infty\), so for all \(n\) and any \(\epsilon>0\), there must be some \(x\in \mathbb{R}\) such that \(|f_n(x)-\cos(1)| \geq \epsilon\). So convergence is not uniform on \(\mathbb{R}\).]
We now consider what happens on \([0,R]\). For fixed \(n\), \[\left|\cos \left(1+\frac{x}{n}\right)-\cos(1)\right|\le \frac{x}{n}\le \frac{R}{n} \quad \hbox{(by the Mean Value Theorem from last year)}.\] So, taking \(s_n= \frac{R}{n}\), we see that \(\{s_n\}\to 0\) as \(n \to\infty\). (By Lemma 5.6 part 1.), this shows that we have uniform convergence (to the constant function \(f(x)=\cos(1)\) on \([0,R]\).
Show that the series
\(\displaystyle
\sum_{k=1}^\infty\frac{2^kz^{2k}}{k^2}\) converges uniformly on
\(%\bar B_{\frac{1}{\sqrt 2}}(0) =
\left\{ z \in \mathbb C: \: |z|\leq\frac{1}{\sqrt{2}}\right\}\),
and deduce that the limit function is continuous on this set.
Let \(f_k(z)=\frac{2^kz^{2k}}{k^2}\). First note that for \(|z|\leq\frac{1}{\sqrt{2}}\) we have \(|z|^{2k}\leq 1/2^k\), so \[|f_k(z)|\quad =\quad\left|\frac{2^kz^{2k}}{k^2}\right| \quad =\quad\frac{2^k|z|^{2k}}{k^2}\quad\leq\quad\frac{1}{k^2}.\] Now \(\sum_{k=1}^{\infty}\frac{1}{k^2}\) is a convergent series, so, by the Weierstrass M-test, the series converges uniformly on \(|z|\leq\frac{1}{\sqrt{2}}\). Since each term of the series is continuous, we see (since uniform limits of continuous functions are continuous) that the limit function \(f(z):= \sum_{k=1}^\infty\frac{2^kz^{2k}}{k^2}\) is continuous. Note that we do not know what \(f(z)\) is, just that it exists and is continuous.
Prove that \(\sum_{n=0}^{\infty} e^{nz}\) converges uniformly on \(\{z \in \mathbb C: \: \mathop{\text{Re}}(z)\le -1\}\), but not on \(\{z \in \mathbb C: \: \mathop{\text{Re}}(z)\le 0\}\).
Let \(f_n(z)=e^{nz}\) and write
\(z=x+iy\), with \(x\le -1\). We have \[|f_n(z)|\: =
\:|e^{nz}|\:=\:|e^{nx}e^{iny}|\:=\:e^{nx} |e^{iny}|\:=\:e^{nx}\:\le\:
e^{-n}\:=\:\left(\frac{1}{e}\right)^n.\] As \(\frac{1}{e}< 1\), the series \(\displaystyle\sum_{n=1}^{\infty}
\left(\frac{1}e\right)^n\) is convergent (it is just a geometric
series). It follows by the Weierstrass M-test that \(\displaystyle\sum_{n=1}^{\infty} e^{nz}\)
converges uniformly on \(\{z \in \mathbb C: \:
\mathop{\text{Re}}(z)\le -1\}\).
The series is not even pointwise convergent on \(\{z \in \mathbb C: \: \mathop{\text{Re}}(z)\le
0\}\); take for example \(z=0\)
for then the partial sums \(\sum_{n=0}^{N}
e^{nz} = N+1\) clearly do not converge. Thus the series does not
converge uniformly on the set \(\{z \in
\mathbb C: \: \mathop{\text{Re}}(z)\le 0\}\).
Let \(R\) satisfy \(0<R<1\). Show that the series \(\displaystyle\sum_{n=1}^{\infty}\frac{z^{n}}{1+z^n}\) converges uniformly on \(%B_R(0)= \{z \in \mathbb C: \: |z|<R \}\). Conclude that the infinite series defines a continuous function on the unit disc \(\mathbb D\).
Let \(f_n(z)=\frac{z^{n}}{1+z^n}\).
Notice that for \(z\) in \(\{z \in \mathbb C: \: |z|<R \}\) we have
\(|z|<1\). So, by the reverse
triangle inequality, \[{\left|1+z^n\right|}
\: = \:{\left|1-(-z^n)\right|} \geq\bigl| \, |1|-{\left|(-z^n)\right|}
\, \bigr| \: = \: \bigl|1-|z|^n\bigr| \: = \: 1-|z|^n \: \geq \:
1-R^n.\] Thus, for \(z\) in
\(\{z \in \mathbb C: \: |z|<R \}\),
we have \[{\left|f_n(z)\right|} \: =
\:{\left|\frac{z^{n}}{1+z^n}\right|} \: \leq \: \frac{|z|^{n}}{1-R^n} \:
< \: \frac{R^{n}}{1-R^n}.\] Let \(M_n=\frac{R^{n}}{1-R^n}\), then (by the
ratio test) the sum \(\sum_{n=1}^{\infty}M_n\) converges. \[\bigl[ \text{\textit{ Details: $L= \lim_{n \to
\infty}\frac{M_{n+1}}{M_n} = \lim_{n \to
\infty}\frac{R^{n+1}/(1-R^{n+1})}{R^{n}/(1-R^{n})} = \lim_{n \to
\infty}\frac{R(1-R^{n})}{1-R^{n+1}} = R < 1$. }}\bigr]\]
Thus, by the Weierstrass M-test, \(\sum_{n=1}^{\infty}\frac{z^{n}}{1+z^n}\)
converges uniformly on \(\{z \in \mathbb C: \:
|z|<R \}\).
To see that the series is continuous on the unit disc, note that for
every point \(z \in \mathbb D\) we can
find \(0<R<1\) such that \(z \in B_R(0)\). Since the series converges
uniformly on \(B_R(0)\) and each
function \(f_n\) is continuous, the
limit function is continuous at \(z\).
[Note that here we have really used uniform convergence - we have found an open set \(B_R(0)\) containing \(z\), on which the series converges uniformly.]
Prove that each of the following series converge uniformly on the corresponding subset of \(\mathbb C\): \[\begin{aligned} (a)&\ \sum_{n=1}^{\infty}\frac{1}{n^2z^{2n}}, &{\rm on}&\quad \{\,z \in \mathbb C: \: {\left|z\right|}\geq 1\,\}.\\ (b)&\ \sum_{n=1}^\infty\sqrt n \, e^{-nz}, &{\rm on}&\quad \{\,z\in \mathbb C: \: \ 0<r\leq \mathop{\text{Re}}(z)\,\}.\\ (c)&\ \sum_{n=1}^\infty\frac{2^n}{z^n+z^{-n}}, &{\rm on}&\quad \left\{\,z\in \mathbb C: \: \ {\left|z\right|}\leq r<\frac{1}{2}\right\}.\\ (d)&\ \sum_{n=1}^\infty2^{-n}\, \cos(nz), &{\rm on}&\quad \{\,z\in \mathbb C: \: \ {\left|\mathop{\text{Im}}(z)\right|}\leq r<\ln 2\, \}. \end{aligned}\]
As in the previous solutions, simply compare each series to \[(a) \: \sum n^{-2}; \quad (b) \: \sum \sqrt n e^{-nr}; \quad (c) \:\ \sum \frac{(2r)^n}{1-r^{2n}}; \quad (d) \:\sum 2^{-n} e^{nr};\] respectively. All converge (we know the series in (a) converges from Analysis I, the series in (b), (c), (d) converge via the ratio test) so the Weierstrass M-test implies uniform convergence of each series.
Given \(0<r<R<\infty\), show that \(\displaystyle\sum_{n=1}^{\infty}\frac{\left(z+\frac{1}{z}\right)^n}{n!}\) converges uniformly on \(r < |z| < R\). Conclude that the infinite series defines a continuous function on \(\mathbb C^{\ast}\).
When \(r < |z|< R\) we have \[\left|z+\frac{1}{z}\right|\leq\left|z\right|+\left|\frac{1}{z}\right|\leq R+\frac{1}{r}.\] Let \(M_n = \frac{\left(R+\frac{1}{r}\right)^n}{n!}\). Then, the sum \(\sum_{n=1}^{\infty}M_n\) converges (by the ratio test). \[\bigl[ \text{\textit{ Details: $L= \lim_{n \to \infty}\frac{M_{n+1}}{M_n} = \lim_{n \to \infty}\frac{\left(R+\frac{1}{r}\right)^{n+1}/(n+1)!}{\left(R+\frac{1}{r}\right)^n/n!} = \lim_{n \to \infty}\frac{R+\frac{1}{r}}{n+1} = 0 < 1$. }}\bigr]\] Thus, the series \(\sum_{n=1}^{\infty}\frac{\left(z+\frac{1}{z}\right)^n}{n!}\) converges uniformly on \(\{z \in \mathbb C: r < |z| < R\}\) by the Weierstrass M-test.
To see that the series is continuous on \(\mathbb C^\ast\), note that for every point \(z_0 \in \mathbb C^\ast\) we can find \(0<r<R<\infty\) such that \(r< |z_0|< R\). Since the series converges uniformly on \(\{z \in \mathbb C: r < |z| < R\}\), this shows that the series converges locally uniformly on \(\mathbb C^*\). Thus by a theorem from class, since all the terms of the series are continuous on \(\mathbb C^*\), the limit function is continuous.
Prove that \(\displaystyle{\sum_{n=1}^\infty \frac{z^n}{1+z^{2n}}}\) converges uniformly on \(|z|< r\), for any \(r<1\). Prove it also converges uniformly on \(|z|\ge R\), for any \(R>1\). Conclude that the infinite series defines a continuous function inside and outside the unit circle. What is the situation on the unit circle?
If \(|z| < r\) for some \(r<1\), then \(|1+z^{2n}|\ge 1-r^{2n}\). Therefore, for \(|z| < r\) \[\left|\frac{z^n}{1+z^{2n}}\right|\le \frac{r^n}{1-r^{2n}}.\] The series \(\sum\frac{r^n}{1-r^{2n}}\) is convergent (by the ratio test). So, by the Weierstrass M-test, the series is uniformly convergent on the set \(\{z \in \mathbb C: \: |z|\le r \}\), for any \(r<1\).
If \(|z|\ge R\) (for \(R>1\)), then \[\left|\frac{z^n}{1+z^{2n}}\right|=\left|\frac{1}{z^n(1/z^{2n}+1)}\right|\le \frac{1}{R^n}\frac{1}{1-1/R^{2n}}=\frac{R^n}{R^{2n}-1}.\] You can use now the same arguments as above to conclude that the series converges uniformly on \(|z|\ge R\), for any \(R>1\).
To see that the series is continuous inside the unit disc, note that for every point \(z_0 \in \mathbb D\) we can find \(0<r<1\) such that \(|z_0|<r\), and so \(z_0 \in B_r(0)\). Since the series converges uniformly on the open ball \(B_r(0)\) and each function \(f_n\) is continuous, the limit function is continuous at \(z\).
[Note that here we have really used uniform convergence - we have found an open set \(B_r(0)\) containing \(z_0\), on which the series converges uniformly.]
Similarly, for \(|z_0|>1\) there is an \(R>1\) such that \(|z_0| > R\) so that \(z_0\) is in the open set \(\{z \in \mathbb C: |z| > R \}\), on which the series converges uniformly. Thus the series is continuous on \(\{z \in \mathbb C: |z| > 1 \}\).
On the unit circle the series does not even converge pointwise. For example, take \(z=1\).