Doubt about step of the proof that $\limsup\limits_{n\to\infty}\left|\sum\limits_{\ell=1}^mz^n_\ell\right|^{1/n}=\max\limits_{1\le\ell\le m}|z_\ell|$ I'm studying some questions of A problem book of complex analysis, Daniel Alpay. The Question is:

Let $z_1$, $\ldots$, $z_m$ be complex different from $0$.
  Show that $$\limsup\limits_{n\to\infty}\left|\sum\limits_{\ell=1}^m{z_\ell}^n\right|^{1/n}=\max\limits_{1\le\ell\le m}|z_\ell|$$

My doubt:
Let $M :=\max\limits_{1\le\ell\le m}|z_\ell|$ and $S_n:=\sum\limits_{\ell=1}^mz_\ell^n$
I prove that $\limsup _{n \rightarrow \infty}|S_n|^{1/n}\leq M$. But I don't saw anything to do for prove that $\limsup |S_n|^{1/n}\geq M$.
I suppose that using contradiction that $\limsup |S_n|^{1/n} = \tilde{M}<M$ and take $z_l = M e^{i2\pi l}$, it'll be done.
Could Someone help me?
 A: We import an elementary result in Number Theory. Lemma: If $(x_1,...,x_k)$ is a  finite sequence of real numbers and  $\epsilon >0$ then there are infinitely many $n\in \Bbb N$ such that  $\max_{1\leq j\leq k} d(nx_j,\Bbb Z)<\epsilon,$ where $d(y,\Bbb Z)=\min \{|y-z|:z\in \Bbb Z\}.$
WLOG, $\;M=1.$ Let $A=\{l:|z_l|=1\}.$ Let $B=\{l:|z_l|<1\}.$ Let $|A|$ be the number of members of $A$. Let $|B|$ be the number of members of $B.$
Note: $|A|\geq 1$ but $B$ could be empty. In what follows, we follow the convention that a sum such as  $\sum_{l\in B}f(l)$ is $0$ if $B$ is empty
Given $\epsilon >0,$ and given $n_0\in \Bbb N:$
(i).   Take $\epsilon'\in (0, \min (1,\epsilon)]$ and take $n_1\geq n_0$ such that $(1-\epsilon')^{1/n_1}>1-\epsilon.$ Then $\forall n \geq n_1\;(\;(1-\epsilon)^{1/n}>1-\epsilon\;).$
(ii). Take $\epsilon''\in (0,\epsilon']$  such that $\forall n\geq n_1\;\forall l\in B\;(\;|z_l|^n<\frac {\epsilon'}{2(1+|B|)}\;).$
(iii). Take $\epsilon'''\in (0,\epsilon'']$ such that $\forall t\in (-\epsilon''',\epsilon''')\;( |1-e^{it}|<\frac {\epsilon'}{2|A|}).$
(iv). By the Lemma, take $n_2\geq n_1$ such that $\forall l\in A\;(d(n_2t_l,\Bbb Z)<\epsilon'''.$ 
(v). Now for $l\in A$ we have $(z_l)^{n_2}=1+\delta_l$ with $|\delta_l|<\frac {\epsilon'}{2|A|}.$  And for $l\in B$ we have $|z_l|^{n_2}<\frac {\epsilon'}{2(1+|B|)}$. 
Hence $$|\sum_{l=1}^m (z_l)^{n_2} |=|\sum_{l\in A}(1-\delta_l) +\sum_{l\in B}(z_l)^{n_2}|   \geq$$   $$\geq |A|-\sum_{l\in A}|\delta_l| - \sum_{l\in B}|z_l|^{n_2}   >$$ $$>|A|-\epsilon'/2 -\epsilon'/2.$$ 
So we have $n_2\geq n_0$  and  $$|\sum_{l=1}^m(z_l)^{n_2}|^{1/n_2}>$$ $$>(|A|-\epsilon')^{1/n_2}\geq$$ $$\geq (1-\epsilon')^{1/n_2} >1-\epsilon =M-\epsilon.$$
A: Let $z_l = r_le^{i\varphi_l}$, $l=1,\ldots,m$, and assume WLOG that $M=|z_1|=r_1$. Put $\phi_l := \varphi_l-\varphi_1$. Then
$$
|S_n|^2 = \left|\sum_{l=1}^mr_l^ne^{in\varphi_l}\right|^2 = \left|\sum_{l=1}^mr_l^ne^{in\phi_l}\right|^2\,\ge\,\left(\sum_{l=1}^mr_l^n\cos(n\phi_l)\right)^2.
$$
Now, as shown by Mark in Given real angles $\phi_1,\ldots,\phi_N$, there exist infinitely many integers $n$ such that $\cos(n\phi_k) > 0$ for all $k$., there are infinitely many $n$ such that $\cos(n\phi_l) > 0$ for all $l=1,\ldots,m$. For these $n$,
$$
|S_n|^2\,\ge\,\left(\sum_{l=1}^mr_l^n\cos(n\phi_l)\right)^2\,\ge\,r_1^{2n}\cos^2(n\phi_1) = M^{2n},
$$
because $\phi_1=0$. Hence, $|S_n|^{1/n}\ge M$ for infinitely many $n\in\Bbb N$, i.e., $\limsup_{n\to\infty}|S_n|^{1/n}\ge M$.
A: I tried to finish with another manner:
Cauchy Hadamard Theorem:
let be the power series $\sum_{n\geq 0} a_n z^n$. Then, the radius of convergence is $R = \frac{1}{\limsup |a_n|^{1/n}}$
consider the function
$g(z) = \sum_{l\in\{1,\ldots,m\}} \frac{1}{1-z_l z} = \sum_{n\geq 0}{(z_1^n +\ldots +z_l^n) z^n}$
The radius of convergence at origin is $ R = \min\frac{1}{|z_l|} = \frac{1}{\max |z_l|}$.
Using the C-H Theorem, it would be done, but It's incomplete.
So, I think about other:
WLOG suppose that
$$ |z_1|\leq |z_2| \leq \ldots \leq |z_k| < |z_{k+1}| =\ldots |z_m| = M$$


*

*We have $m-k$ z's with maximum modulus. Applying Triangle Inequality, we have


$$ \Bigg|\sum_{l=0}^{m}z^n_l \Bigg|^{1/n} \leq \Bigg(\sum_{l=0}^{m}|z|^n_l\Bigg) \leq (m M^n)^{1/n}\Rightarrow \Bigg|\sum_{l=0}^{m}z^n_l \Bigg|^{1/n} \leq m^{1/n} M$$
So
$$ \lim_{n\rightarrow \infty} m^{1/n} M = M \Rightarrow \limsup_{n\rightarrow\infty} \Bigg|\sum_{l=0}^{m}z^n_l \Bigg|^{1/n} \leq M$$


*Divide the expression by $z^n_m = M^n e^{in\alpha_m}$:


$$\Bigg(\sum_{l=0}^{m}z^n_l \Bigg) = z^n_m\Bigg(\sum_{l=0}^{m}\Big(\frac{z_l}{z_m} \Big)^n\Bigg) = z^n_m\Bigg(\sum_{l=0}^{k}\Big(\frac{z_l}{z_m} \Big)^n+ \sum_{l=k+1}^{m}\Big(\frac{z_l}{z_m} \Big)^n\Bigg) = z^n_m\Bigg(\sum_{l=0}^{k}\Big(\frac{z_l}{z_m} \Big)^n+ \sum_{j=k+1}^{m}e^{in\theta_j} \Bigg)\Rightarrow$$
$$\Rightarrow \Bigg|\sum_{l=0}^{m}z^n_l \Bigg|^{1/n} = |z_m|\Bigg|\sum_{l=0}^{k}\Big(\frac{z_l}{z_m} \Big)^n+ \sum_{j=k+1}^{m}e^{in\theta_j} \Bigg|^{1/n}$$
So, we have
$$\limsup_{n\rightarrow \infty}  \Bigg|\sum_{l=0}^{m}z^n_l \Bigg|^{1/n} = M \limsup_{n\rightarrow \infty} \Bigg|\sum_{l=0}^{k}\Big(\frac{z_l}{z_m} \Big)^n+ \sum_{j=k+1}^{m}e^{in\theta_j} \Bigg|^{1/n}$$


*We need to prove that
$$\limsup_{n\rightarrow \infty} \Bigg|\sum_{l=0}^{k}\Big(\frac{z_l}{z_m} \Big)^n+ \sum_{j=k+1}^{m}e^{in\theta_j} \Bigg|^{1/n} = 1$$



Fact: The orbit of irrational rotation on the circle are dense in the unit circle.



*Consider this one 


$$ \Bigg|\sum_{l=0}^{k}\Big(\frac{z_l}{z_m} \Big)^n+ \sum_{j=k+1}^{m}e^{in\theta_j} \Bigg|$$
We know that 
$$\lim_{n\rightarrow \infty} \Big(\frac{z_l}{z_m} \Big)^n = 0$$
But, the second therm we could have rational angles $p_j/q_j$ and irrationals $\alpha_j$.


*WLofG we need to prove that


$$\limsup_{n\rightarrow \infty}\Bigg|\sum_{l=k+1}^{k'}e^{in\alpha_j}+ \sum_{j=k'+1}^{m}e^{inp_i/q_i} \Bigg| = m-k$$
For rationals, take a subsequence of $N_n = n (2 \prod_j q_j)$. This sequence will land all angles at zero.


*1 is a accumulation point. using the fact above(that irrational rotation is dense in unit circle), $e^{n\alpha_j}$ never repeat your values.


Using the Diophantine approximation for $\alpha_j \in \mathbb{R}\backslash\mathbb{Q}$ this numbers become near to rationals. with this, we can approximate $e^{i\alpha_j}$ for any $e^{ip/q}$. We have an accumulation point  for each $n\alpha_j$, that is, for any $\epsilon >0$ there are two iterates $m>n$ such that $|e^{im\alpha_j}-e^{in \alpha_j}|<\epsilon$, so $|e^{i(m-n)\alpha_j}-1|<\epsilon$.
So, this number is $\epsilon$ close to $1$. We are done.
