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 $\limsup |S_n|^{1/n} = \tilde{M}&lt;M$ and take $z_l = M e^{i2\pi l}$, it'll be done.

Could Someone help me?
We import an elementary result in Number Theory. Lemma: If $(x_1,...,x_k)$ is a  finite sequence of real numbers and  $\epsilon &gt;0$ then there are infinitely many $n\in \Bbb N$ such that  $\max_{1\leq j\leq k} d(nx_j,\Bbb Z)&lt;\epsilon,$ where $d(y,\Bbb Z)=\min \{|y-z|:z\in \Bbb Z\}.$

WLOG, $\;M=1.$ Let $A=\{l:|z_l|=1\}.$ Let $B=\{l:|z_l|&lt;1\}.$ 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 $\epsilon &gt;0,$ and given $n_0\in \Bbb N:$

(i).   Take $\epsilon'\in (0, \min (1,\epsilon)]$ and take $n_1\geq n_0$ such that $(1-\epsilon')^{1/n_1}&gt;1-\epsilon.$ Then $\forall n \geq n_1\;(\;(1-\epsilon)^{1/n}&gt;1-\epsilon\;).$

(ii). Take $\epsilon''\in (0,\epsilon']$  such that $\forall n\geq n_1\;\forall l\in B\;(\;|z_l|^n&lt;\frac {\epsilon'}{2(1+|B|)}\;).$

(iii). Take $\epsilon'''\in (0,\epsilon'']$ such that $\forall t\in (-\epsilon''',\epsilon''')\;( |1-e^{it}|&lt;\frac {\epsilon'}{2|A|}).$

(iv). By the Lemma, take $n_2\geq n_1$ such that $\forall l\in A\;(d(n_2t_l,\Bbb Z)&lt;\epsilon'''.$ 

(v). Now for $l\in A$ we have $(z_l)^{n_2}=1+\delta_l$ with $|\delta_l|&lt;\frac {\epsilon'}{2|A|}.$  And for $l\in B$ we have $|z_l|^{n_2}&lt;\frac {\epsilon'}{2(1+|B|)}$. 

Hence $$|\sum_{l=1}^m (z_l)^{n_2} |=|\sum_{l\in A}(1-\delta_l) +\sum_{l\in B}(z_l)^{n_2}|   \geq$$   $$\geq |A|-\sum_{l\in A}|\delta_l| - \sum_{l\in B}|z_l|^{n_2}   &gt;$$ $$&gt;|A|-\epsilon'/2 -\epsilon'/2.$$ 

So we have $n_2\geq n_0$  and  $$|\sum_{l=1}^m(z_l)^{n_2}|^{1/n_2}&gt;$$ $$&gt;(|A|-\epsilon')^{1/n_2}\geq$$ $$\geq (1-\epsilon')^{1/n_2} &gt;1-\epsilon =M-\epsilon.$$
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

$$ |z_1|\leq |z_2| \leq \ldots \leq |z_k| &lt; |z_{k+1}| =\ldots |z_m| = M$$


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 $\epsilon &gt;0$ there are two iterates $m&gt;n$ such that $|e^{im\alpha_j}-e^{in \alpha_j}|&lt;\epsilon$, so $|e^{i(m-n)\alpha_j}-1|&lt;\epsilon$.

So, this number is $\epsilon$ close to $1$. We are done.