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I'm learning "Complex Analysis", section Series and Convergence, and I got stuck on this problem (actually, just a small part of this problem):

Find all the values of $z$ which makes this series normal converges and uniformly converges in complex plane $\mathbb C$: $$\sum_{n=1}^\infty{(-1)^n}{z^n \over n}$$

Using Cauchy-Hadamard formula, I know that the convergence radius is 1, so this series will converge normally in $B(0, 1)$. But what about the point on boundary of $B(0,1)$. For example, if we choose $z=1$, then this series will converge, but if $z=-1$, then this series will not converge. So what about other values which satisfies $|z| = 1$? And can you give me some hint to work on uniformly convergence? Thanks

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  • $\begingroup$ What do you want to know about uniform convergence ? Uniform convergence is not a local notion; and you already know that your power series converges uniformly on any compact subset of $B(0,1)$, but does not converge uniformly over $\bar B(0,1)$ (uniform convergence $\implies$ pointwise convergence). $\endgroup$ – Gabriel Romon Mar 13 '16 at 18:01
  • $\begingroup$ Regarding pointwise convergence on the unit circle, you can use Dirichlet test with the series $\sum \frac{e^{i\theta n}}{n}$. $\endgroup$ – Gabriel Romon Mar 13 '16 at 18:05
  • $\begingroup$ @LeGrandDODOM: I thought the convergence we found using convergence radius is normally convergence, not uniformly convergence. That's what I found in my book. Can you make it clearer, or cite some source so I can read? Thanks $\endgroup$ – le duc quang Mar 14 '16 at 2:23
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Normal convergence is great, because it implies uniform convergence, in some way. Let me restate some definitions first.

A series $\sum_n f_n$ of functions $f_n:X\to \mathbb C$ is said to be normally convergent in $X$ if each $x\in X$ has a neighborhood $U$ such that $\sum_n |f_n|_U<\infty$ where $|f_n|_U= \sup_U |f_n|$.

A series $\sum_n f_n$ of functions $f_n:X\to \mathbb C$ is said to be locally uniformly convergent in $X$ if every $x\in X$ lies in a neighborhood $U_x$ in which the series $\sum_n f_n$ converges uniformly.

A series $\sum_n f_n$ of functions $f_n:X\to \mathbb C$ is said to be compactly convergent in $X$ if it converges uniformly on every compact subset of $X$.

Claim: normally convergent $\implies$ locally uniformly convergent

Proof:

Fix some $x\in X$. Since $\sum_n f_n$ is normally convergent, there is some neighborhood $U$ of $x$ such that $\sum_n |f_n|_U<\infty$. Let us prove that $\sum_n f_n$ converges uniformly over $U$.

Let $\epsilon >0$. Since the series $\sum_n |f_n|_U<\infty$ converges, there exists some $N\in \mathbb N$ such that $n\geq N\implies \sum_{k=n}^\infty |f_k|_U \leq \epsilon$. Consider some $y\in U$ and $n\geq N$. Then $\left| \sum_{k=n}^\infty f_k(y) \right|\leq \sum_{k=n}^\infty |f_k(y)| \leq \sum_{k=n}^\infty |f_k|_U \leq \epsilon$

This proves uniform convergence over $U$ and we're done.

Claim: locally uniformly convergent $\implies$ compactly convergent

Proof: Let $K$ be a compact subset of $X$. Every $x\in X$ has a neighborhood $U_x$ on which the series $\sum_n f_n$ converges uniformly. Hence $K\subset \cup_{x\in K} U_x$. Since $K$ is compact, there exists a finite subcover of $K$: $K\subset U_{x_1}\cup\ldots\cup U_{x_N}$. It's easy to check that $\sum_n f_n$ converges uniformly on $U_{x_1}\cup\ldots\cup U_{x_N}$, and a fortiori on K.


You proved that $\sum_{n\geq1}{(-1)^n}{z^n \over n}$ converges normally on $B(0,1)$. By the claims above, it converges compactly on $B(0,1)$ (that is to say, it converges uniformly on each compact subset of $B(0,1)$).


Regarding pointwise convergence on the unit circle, use Dirichlet's test to prove that $\sum \frac{e^{i\theta n}}{n}$ converges whenever $\theta \neq 2k\pi$.


Claim: $\sum_{n\geq1}{(-1)^n}{z^n \over n}$ does not converge uniformly on $B(0,1)$.

Proof: Suppose for contradiction that $\sum_{n\geq1}{(-1)^n}{z^n \over n}$ converges uniformly on $B(O,1)$. By Cauchy's uniform convergence criterion, for any $\epsilon >0$, there exists $N$ such that $$\forall z\in B(0,1), \forall p, \forall n, \;\; p> n\geq N\implies \left| \sum_{k=n+1}^p{(-1)^n}{z^n \over n}\right|\leq \epsilon$$

Fixing $p$ and $n$ and letting $z$ tend to $-1$ yields $$\forall p, \forall n, \;\; p> n\geq N\implies \left| \sum_{k=n+1}^p{1 \over n}\right|\leq \epsilon$$

Using Cauchy convergence criterion for series, this implies that the harmonic series converges, a contradiction.


To put an end to OP's question, it's worth noticing that the series of functions converges on any circular arc that doesn't intersect $-1$. See here for a proof.

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  • $\begingroup$ Correctly. Thanks, but I wonder in the disk $B(0,1)$, does the series converge uniformly? You proved that in this disk, the series locally uniformly converges, but it does not imply the series uniformly converges in the disk $B(0,1)$, right? $\endgroup$ – le duc quang Mar 14 '16 at 14:26
  • $\begingroup$ @leducquang I made an edit to answer your question. $\endgroup$ – Gabriel Romon Mar 14 '16 at 15:12
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I would like to bring some precisions.

First of all, I take the opposite of your series, which means that the function we consider is

$$\sum_{n=1}^{\infty}(-1)^{n+1} {z^n \over n}=\ln(1+z) \ \ \ (1)$$

this relationship being obtained by integration of the series:

$$\sum_{n=1}^{\infty} (-1)^{n+1} z^{n-1}=\dfrac{1}{1+z}$$

Thus

  • taking $z=1$ in (1), you may know that the real series $\sum_{n=1}^{\infty}(-1)^{n+1}\dfrac{1}{n} \ \ \ \ \ \ $ (harmonic alternated series) is convergent (to $\ln 2$).

  • taking $z=-1$ in (1), on one hand, it would surprising that $\ln 0$ has a meaning. Moreover, we are faced with the harmonic series, which is known to be divergent.

  • taking any other value of $z$ on $C(0,1)$, there is convergence, but not absolute convergence (because of the divergent character of harmonic series)..

In order to illustrate this last case, consider value $z=-i$; let us see which meaning should be attributed to

$$\ln(1+i)=i-\frac{i^2}{2}+\frac{i^3}{3}+ \cdots$$

The finite sums of this series draw a converging spiral around a limit point $\approx 0.347+i0.785$, the true value being $\frac12 \ln 2 + i \frac{\pi}{4}$. This shouldn't be astonishing because of relationship $1+i=\sqrt{2}e^{i\pi/4}$, from which $\ln(1+i)=\ln \sqrt{2} + \ \ln e^{i\pi/4}$... (Yes, I agree, I hide a certain number of theoretical issues...)

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Finally, I advise you to have a look at the interesting article https://en.wikipedia.org/wiki/Radius_of_convergence where the convergence of $-\ln(1-z)$ is considered.

Remark for (1): if you meet the still widespread notation $\log(1+z)$ instead of (1), it is the same function (not to be confused with the base-10 logarithm).

Edit: As you hadn't mentionned "log" function, I thought that you didn't noticed it. So now I understand that it was only an example of what you are looking for. The key concept is that the disk of convergence can be thought as the limit position of a disk with growing radius that you "inflate" till you reach the first singularity (here, with $\log(z+1)$, it is $z=-1$). If, by chance there are other singularities on the very same disk, it will be the other points of non convergence you are looking for. But usually, there are no other singularity on this boundary circle. This is the case here due to the "cut" of function $\ln(1+z)$ that only forbids points in $(-\infty,-1]$ on the negative real axis.

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  • $\begingroup$ I understand what you mean, but how can we prove formally that for every $z$ different from -1, the series converges? What you said seems just the institution and representation. $\endgroup$ – le duc quang Mar 14 '16 at 2:15
  • $\begingroup$ See the "Edit" on my previous answer $\endgroup$ – Jean Marie Mar 14 '16 at 8:38
  • $\begingroup$ I see what you mean. So maybe my teacher didn't say the theory completely. From what I learnt, we can only apply the equation (1) only when $|z| < R$. When $|z| = R$, then we don't have that equation anymore. So you mean, even when $|z| = R$, we can still use that equation in domain of the right-hand function (here the function $\log$), right? $\endgroup$ – le duc quang Mar 14 '16 at 14:17
  • $\begingroup$ Yes, more or less. An essential step in the understanding of the domains of validity of Taylor expansions (and the way they are compatible) is the principle of "analytic continuation". See en.wikipedia.org/wiki/Analytic_continuation $\endgroup$ – Jean Marie Mar 15 '16 at 11:02

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