# question about limit and series

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 $\forall1\le j\le m\;$ , $\;a_j=1$ .

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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

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 $1$.
@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