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Here's a problem I'm having trouble with:

Show that if $\sum u_k$ converges for $u_k\in\Bbb C$, and $|\arg(u_k)|\leq c<\pi/2$ for all $k$, then $\sum |u_k|$ converges too.

All I have after looking at it for an hour is that I kind of believe it could be true (I didn't at first), but I don't really know where to start. Could you give me a hint? I know of course that $$u_k=|u_k|(\cos(\arg(u_k))+i\sin(\arg(u_k))),$$ but I don't have any ideas. I've tried looking at it geometrically too, but I just don't see anything.

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2 Answers 2

up vote 3 down vote accepted

Let $S = \{z\in \mathbb{C} : |\arg(z)|\leq c\}$. This is a sector of the right half plane. We can make two immediate observations about points in this sector:

  • If $z\in S$, then the real part of $z$ is nonnegative.
  • There is a constant $A>0$ such that if $z = x + iy\in S$, then $|y|\leq Ax$. The constant $A$ is the absolute value of the slope of the bounding lines of the sector $S$.

Using the second bullet point, we see that if $z = x + iy\in S$, then $$|z|^2 = x^2 + y^2 \leq (1+A^2)x^2,$$ and hence $$|z|\leq \sqrt{1+A^2}x = \sqrt{1+A^2} Re(z).$$

We can then apply this to the $u_k$, to derive that $$\sum_{k=1}^N|u_k|\leq \sqrt{1+A^2}\sum_{k=1}^N Re(u_k)\leq \sqrt{1+A^2} Re\sum_{k=1}^Nu_k\leq \sqrt{1 + A^2}\left|\sum_{k=1}^N u_k\right|.$$ We know that $|\sum_{k=1}^N u_k|$ converges as $N\to \infty$, since $\sum u_k$ converges. Thus the right hand side is bounded. It follows that the sequence $\sum_{k=1}^N |u_k|$ is bounded in $N$. But this sum increases as $N\to \infty$, and all bounded increasing sequences converge. Thus $\sum |u_k|$ converges.

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I think I have a counter example. Take $u_k=\frac{1}{k^2}+i\frac{(-1)^k}{k}$. Since $\Re{(u_k)}>0$ for all $k$, we satisfy $\vert \arg(u_k)\vert<\pi/2$. We also have that $\sum u_k$ converges. Yet $\vert u_k\vert=\sqrt{1/k^4+1/k^2}>\sqrt{1/k^2}=1/k$ and hence $\sum\vert u_k\vert$ diverges.

Edit: This is a faulty counter-example, since it doesn't satisfy the condition $\vert\arg(u_k)\vert\leq c<\pi/2$. In fact, $\vert\arg(u_k)\vert\rightarrow\pi/2$ as $k\rightarrow\infty$. @froggie's proof is correct.

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The condition is stronger than just $|\arg(u_k)|<\pi/2$, and your sequence doesn't satisfy the stronger condition. –  Jonas Meyer Dec 16 '12 at 0:11
    
Excellent! I knew something didn't quite seem right. –  icurays1 Dec 16 '12 at 0:14
1  
+1 for an excellent example of why $|\arg u_k|<\pi/2$ doesn't suffice. –  Jonas Meyer Dec 16 '12 at 0:17

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