Why $\lim_{n\to+\infty}\bigg(\bigg\|\sum_{f\in F(n,d)} A_{f}^* A_{f}\bigg\|^{\frac{1}{2n}} \bigg)\;\text{exists}?$ Let $E$ be a complex Hilbert space and $\mathcal{L}(E)$ be the algebra of all bounded linear operators on $E$.

For $A= (A_1,\cdots,A_d)\in\mathcal{L}(E)^d$  (not necessary to be commuting). Why
  $$\lim_{n\to+\infty}\bigg(\bigg\|\sum_{f\in F(n,d)} A_{f}^* A_{f}\bigg\|^{\frac{1}{2n}} \bigg)\;\text{exists}?,$$
  where $F(n,d)$ denotes the set of all functions from $\{1,\cdots,n\}$ into  $\{1,\cdots,d\}$ and $A_f:=A_{f(1)}\cdots A_{f(n)}$, for $f\in F(n,d)$.

I try to apply Fekete's lemma:
For every subadditive sequence $(a_n)_{n\in \mathbb{N}^*}$, the limit $\displaystyle \lim _{n\to \infty }\frac{a_n}{n}$ exists and is equal to $\displaystyle \inf_{n\in \mathbb{N}^*}\frac{a_n}{n}$. (The limit may be ${\displaystyle -\infty }$).
But I'm facing difficulties in the choice of the sequence $(a_n)_{n\in \mathbb{N}^*}$.
Thank you for your help.
 A: Short answer: because it's an instance of the spectral radius formula. 
Details. Let's start with the following inequality on operator norms. Given a finite family $A:=(A_j)_{j\in J}$ of operators $A_j \in\mathcal{L}(H)$, and a symmetric operator $L\in \mathcal{L}_{sym}(H)$, one has
$$\big\|\sum_{j\in J}A_j^*LA_j\big\|\le \big\|\sum_{j\in J}A_j^*A_j\big\|\ \|L\|\ ,$$
with an obvious equality, incidentally, for $L=I$. Indeed, for any $x\in H$, 
$$ \big(\sum_{j\in J}A_j^*LA_jx, \ x\big) =  \sum_{j\in J}(A_j^*LA_jx,x) =  \sum_{j\in J}(LA_jx,A_jx)\le   \sum_{j\in J}\|L\|(A_jx,A_jx)$$
$$=  \|L\| \sum_{j\in J}(A_j^*A_jx,x)\le\|L\|\ \big\|\sum_{j\in J}A_j^*A_j\big\|\ \|x\|^2,$$
and since the same holds for $-L$ the claimed inequality between operator norms follows.
We can interpret the latter in the following simple way: the bounded linear operator ${\bf T} \in\mathcal{L} \big(\mathcal{L}_{sym}(H)\big)$ defined by ${\bf T} L=\sum_{j\in J}A^*_jLA_j\in \mathcal{L}_{sym}(H)$ for any $L\in \mathcal{L}_{sym}(H)$ has operator 
norm $\|{\bf T} \|=\|\sum_{j\in J}A_j^*A_j\big\|$. 
Also note that ${\bf T}^n$ is of the same nature of ${\bf T}$, corresponding to a larger family, namely $(A_f)_{f\in J^n}$:
$${\bf T}^nL=\sum_{f\in J^n} A_f^*LA_f    $$
so that in particular $\|{\bf T}^n\|=\|\sum_{f\in J^n} A_f^*A_f\|$, and the limit you wrote is (the square root of) the spectral radius of ${\bf T}$, namely $$\rho({\bf T})=\lim_{n\to \infty}\|{\bf T}^n\|^{1/n}=\inf_{n\in\mathbb{N}}\|{\bf T}^n\|^{1/n}.$$
A: A simple calculation shows that 
$$\sum_{f\in F(n+m,d)}A_f^*A_f=\sum_{g\in F(m,d)} A_g^*\left(\sum_{f\in F(n,d)}A_f^*A_f\right)A_g$$
Let $a_n:=\left\|\sum_{f\in F(n,d)}A_f^*A_f\right\|$. By use of the above equality we have 
$$0\leq a_{m+n}\leq a_ma_n$$
so $\lim_n a_n^{\frac 1n}$ exists.
