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consider following hypotheses

  • $ m\in\mathbb N$
  • $c\in \mathbb C\,$ ,$\, \; a_j\in \mathbb C$
  • $a_j\in \mathbb C\;$ , $\;|a_j|=1,\;\forall\;1\le j\le m$

if $$\lim\limits_{n\to+\infty}\sum_{j=1}^{m}a^n_j=c$$

then how to prove $ c=m$ and $\forall1\le j\le m\;$ , $\;a_j=1$ .

Thanks in advance

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Of course $c=m$ follows easily from "$\forall1\le j\le m:a_j=1$". I feel like that is the hard part though. –  alex.jordan Mar 13 '13 at 20:03
    
I don't see that the limit necessarily converges. For the case $a_1=1, a_2=-1$ it seems like $\sum_{j=1}^2 a_j^n$ is 2 when $n$ is even and 0 otherwise. Are you missing some assumption? –  Peder Mar 13 '13 at 20:07
    
@Peder I think it is implied that the limit exists. I don't mean it necessarily is a consequence of the hypothesis, I just mean the reader is supposed to assume the limit exists. –  Git Gud Mar 13 '13 at 20:08
    
"contest-math"? What contest? –  Gerry Myerson Mar 13 '13 at 23:23
    
@Gerry Myerson:this question is from Iran competition math –  Maisam Hedyelloo Mar 14 '13 at 3:02

1 Answer 1

Here's an outline: Each $a_j$ is of the form $\exp(\pi i b_j)$. Each $b_j$ can be approximated by a rational number. Then high powers of $a_j$ can be seen to get arbitrarily close to $1$ cyclically, using least common denominators of the $b_j$. This makes the total close to $m$.

And if $k$ of the $a_j$ are not equal to $1$, then high powers of those $a_j$ will cyclically have negative real part, making the total have real part $\le m-k$. Since the limit exists, the only resolution to this is that $k=0$ and they are all equal to $1$.

Again, this is just an outline. All the bits about cycling and closeness would need to be formalized for a solid argument.

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@GitGud Yes, $b_j$ is defined in $[0,1)$ by $a_j=\exp(\pi i b_j)$. –  alex.jordan Mar 13 '13 at 20:20
    
@alexjordan I had missed that, sorry. I skipped over when I read $\exp$. –  Git Gud Mar 13 '13 at 20:21
    
Oops, I guess I meant $[0,2)$ given my parameterization. –  alex.jordan Mar 13 '13 at 21:00
    
If $b_j=2/3$, then high powers of $a_j$ do not get arbitrarily close to $-1$. –  Gerry Myerson Mar 13 '13 at 23:28
    
Thanks @Gerry! Hmm...maybe we can just arrange for high powers to have negative real part. –  alex.jordan Mar 14 '13 at 0:10

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