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Let $\{z_n\}$ be a sequence of complex numbers, and suppose $\sum\limits_{k=0}^\infty z_k$ converges. Suppose $\exists \delta>0$, with $\delta < \pi/2$ such that $|\arg(z_k)|\le \delta$ for all $k\ge 0$. Prove that $\sum\limits_{k=0}^\infty z_k$ converges absolutely.

Proof:

Let $Z:= \sum\limits_{k=0}^\infty z_k$. Since $\sum\limits_{k=0}^\infty z_k$ converges, $z_k$ also converges, so let $z:=\lim\limits_{k\to \infty}z_k$. This implies that $\forall\varepsilon > 0, \exists N_1>0$ such that $|z_k-z|<\varepsilon$ whenever $k>N$. Also, $\exists N2$ such that $k>N_2$ implies $\left| \sum\limits_{k=0}^\infty z_k -Z\right|<\varepsilon$. Let $N=\max\{N_1, N_2\}$. Let $z_k = r_k e^{i\theta_k}$ and $\sum\limits_{k=0}^N z_k=se^{i\gamma_k}$, then

$$\left|\sum\limits_{k=0}^n |r_k e^{i\theta_k}|-|se^{i\gamma}|\right|...$$

But what next? What is so special about the absolute value of the argument of $z_k$ being less than $\pi/2$? Please help.

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    Note that this is **not** true if $\delta = \pi/2$. For example take $z_k = 2^{-k}+(-1)^ki$. The fact that $\delta$ is **strictly** less than $\pi/2$ is absolutely essential for this to be true. – 2017-02-17
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    Sorry, what should I note about $2^{-k}+(-1)^k i$? I see that its argument is $\pi/2$, but what follows from this? – 2017-02-17
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    It is true that $\sum z_k$ converges conditionally, but not absolutely, even though $|\arg z_k| < \pi/2$ for all $k$. – 2017-02-17

1 Answers 1

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Since $z \mapsto Re(z)$ is continuous it follows that $\sum Re(z_k)$ also converges. Note that $Re(z_k)>0$ for all $k$, and thus $\sum Re(z_k)$ converges absolutely.

Now, using the given information we see that $|Im(z_k)/Re(z_k)| \leq \tan \delta$, therefore $|Im(z_k)| \leq \tan \delta |Re(z_k)|$. So by the comparison test $\sum Im (z_k)$ also converges absolutely.

Now using the fact that $|z_k| \leq |Im(z_k)|+|Re(z_k)|$ the claim follows easily.

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    Can you please clarify why $|Im(z_k)/Re(z_k)|\le \tan \delta$? – 2017-02-17
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    @sequence Because $Im(z)/Re(z) = \tan \arg z$. This is nothing more than the formula that we learned in high school: The tangent equals opposite over adjacent (soh-cah-toa) – 2017-02-17
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    It should be true that $|Im(z_k)/Re(z_k)| \le \delta$, correct? For if $z=x+iy$, then $\arg(z_k) = tan(y/x)\le tan\delta$, so $|y/x| \le \delta$. @Shalop – 2017-02-17
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    @sequence $\text{Re}(z)=|z|\cos(\arg(z))$ and $\text{Ie}(z)=|z|\sin(\arg(z))$. Hence, $\frac{\text{Im}(z)}{\text{Re}(z)}=\tan(\arg(z))$ – 2017-02-17
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    @Dr.MV Yes, I understand. What I mean is that it appears that $|Im(z_k)/Re(z_k)|$ is also less than $\delta$. – 2017-02-17
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    @sequence No it is not true that $\arg(z_k) = \tan(y/x)$. Rather it is true that $\arg(z_k) = \tan^{-1}(y/x)$. In fact $\tan(\delta)\to \infty$ as $\delta \to \pi/2$. – 2017-02-17
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    @Shalop Oops, yes. But the above still appears to hold. That is, $\arg(z_k)=tan^{-1}(y/x)\le \tan^{-1}(\delta) \implies y/x \le \delta$. – 2017-02-17
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    @sequence No we are given in the problem that $\arg(z_k) \leq \delta$. therefore $\tan^{-1}(y/x) = \arg(z_k) \leq \delta$ and thus $y/x \leq \tan(\delta)$. – 2017-02-17
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    @Shalop I agree. Nevertheless, $\arctan(\theta)$ is an increasing function for $\theta\ge0$, and for $0\le \delta\le \pi/2$, $\arg(z_k) = \arctan(y/x) \le \arctan(\delta)$. Correct? Hence, $y/z\le \delta$. Please let me know where I'm not reasoning correctly if this is not the case. – 2017-02-17
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    @sequence No that's not correct because it is not true that $y/x <\delta$. – 2017-02-17
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    @Shalop Can you please clarify what theorem you were using when you argued that since $f: z\to \Re\{z\}$ is continuous and $\sum z_k$ is convergent, then $\sum f(z_k)$ is also convergent? – 2017-02-17
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    Let $s_n$ denote the $n^{th}$ partial sum of $z_k$, so $s_n = \sum_1^n z_k$. Then $Re(s_n) = \sum_1^n Re(z_k)$. If $s_n \to s$, then $Re(s_n) \to Re(s)$ by continuity. Hence $\sum_1^{\infty} Re(z_k)$ is convergent and the sum is $Re(s)$. – 2017-02-17