Statement about limits superior 

For $i=1,\ldots,k$ let $(a_n^{(i)})_{n\in\mathbb{N}}$ be a sequence of positive real numbers. Then
    $$
\limsup_{n\to\infty}\frac{1}{n}\log(a_n^{(1)}+\ldots+a_n^{(k)})=\max_{1\leq i\leq k}\left\{\limsup_{n\to\infty}\frac{1}{n}\log a_n^{(i)}\right\}.
$$


I have some problems to understand the following proof.
"$\le$" is clear since for every $i\in\left\{1,...,k\right\}$ we have $a_n^{(1)}+...+a_n^{(k)}\geq a_n^{(i)}$ and hence the left-hand side of the equality is greater than or equal the right-hand side.
But "$\geq$" males me some problems:
Let $R$ denote the right-hand side of the equality. If $R=\infty$, then there remains nothing to show. If $R$ is finite, then for every $i\in\left\{1,...,k\right\}$ one has $\limsup_{n\to\infty}\frac{1}{n}\log a_n^{(i)}\leq R$.


So if $\varepsilon >0$, then
    $$
a_n^{(i)}\leq\exp(n(R+\varepsilon))~\text{ for almost all }n.~~~~~(*)
$$
    Consequently,
    $$
a_n^{(1)}+...+a_n^{(k)}\leq n\cdot\exp(n(R+\varepsilon)).~~~~(**)
$$
    It follows that
    $$
\frac{1}{n}\log(a_n^{(1)}+...+a_n^{(k)})\leq \frac{1}{n}\log n+R+\varepsilon~\text{for almost all }n.~~~(***)
$$
    Thus, the left-hand side of the equality above is less than or equal to $R+\varepsilon$. Since this holds for every $\varepsilon >0$, the desired result follows.


I do not understand the statements $(*)$, $(**)$ and $(***)$, especially the "for almost all n".
Could you pls explain them to me?
 A: The "for almost all $n$" means "there is an $N\in \mathbb{N}$ such that for all $n \geqslant N$". A property holds "for almost all $n$" if it holds for all but finitely many $n$, so it holds for all "sufficiently large" $n$.
Here, by definition of $\limsup$, for every $\varepsilon > 0$, there is, for every $i$ an $N_i(\varepsilon)$ such that
$$\frac{1}{m}\log a_m^{(i)} < \varepsilon + \limsup_{n\to\infty}\frac{1}{n}\log a_n^{(i)}\tag{1}$$
holds for all $m \geqslant N_i(\varepsilon)$. By definition of $R$, we thus have
$$\frac{1}{m}\log a_m^{(i)} < R + \varepsilon\tag{2}$$
for $m \geqslant N_i(\varepsilon)$. Let $N(\varepsilon) = \max\limits_{1\leqslant i \leqslant k} N_i(\varepsilon)$. Then for $m \geqslant N(\varepsilon)$, the inequality $(2)$ holds for all $i$. And thus, by multiplication and exponentiation we reach
$$n \geqslant N(\varepsilon) \implies a_n^{(i)} < \exp\bigl(n\cdot (R+\varepsilon)\bigr)\tag{$\ast$}$$
for $1 \leqslant i \leqslant k$. Summing the inequalities $(\ast)$ for $1 \leqslant i \leqslant k$, we obtain
$$n\geqslant N(\varepsilon) \implies a_n^{(1)} + \dotsc + a_n^{(k)} < k\cdot \exp\bigl(n(R+\varepsilon)\bigr).\tag{$\ast\!\ast$}$$
In your question, the factor in front of the exponential is $n$ rather than $k$. For $n > k$, that is a weaker inequality, hence implied by $(\ast\ast)$ here. For $n < k$, it might be wrong, but since we are interested in the behaviour as $n \to \infty$ and can always choose $N(\varepsilon) \geqslant k$, that's not something to worry about. However, it is conceptually irritating to choose a factor of $n$ there rather than the fixed $k$, since on the left hand side we have a sum of $k$ terms, each bounded above by the same expression. Hence it is more natural to let the right hand side be $k$ times the bound for each of the terms.
Now, taking logarithms in $(\ast\ast)$, where we denote the factor by $F$, which could be either of $k$ or $n$ [for sufficiently large $n$], we obtain
$$\log \bigl(a_n^{(1)} + \dotsc + a_n^{(k)}\bigr) < n(R+\varepsilon) + \log F.$$
Then we can divide by $n$ to obtain
$$\frac{1}{n}\log \bigl(a_n^{(1)} + \dotsc + a_n^{(k)}\bigr) < R + \varepsilon + \frac{\log F}{n} \tag{$\ast\!\ast\!\ast$}$$
for all $n \geqslant N(\varepsilon)$. Since $\frac{\log F}{n} \to 0$, that shows
$$\limsup_{n\to\infty} \frac{1}{n}\log \bigl(a_n^{(1)} + \dotsc + a_n^{(k)}\bigr) \leqslant R + \varepsilon.$$
Since $\varepsilon > 0$ was arbitrary,
$$\limsup_{n\to\infty} \frac{1}{n}\log \bigl(a_n^{(1)} + \dotsc + a_n^{(k)}\bigr) \leqslant R$$
follows.
