As Qiaochu said, the statement $\lim\limits_{n \to \infty}{a_n}=2^{\frac{k}{k-1}}$ requires further justification. However, it can be proven as follows:
Intuitive Explanation: We will show that there are three possible behaviors for the sequence $a_n,n\in\mathbb{N}$. First, eventually all terms will be less than $2^{\frac{k}{k-1}}$ and the sequence will be increasing. Second, eventually all terms will be greater than $2^{\frac{k}{k-1}}$ and the sequence will be increasing. Either of these means the limit exists, and using lemmas 1 and 3 we see that it cannot be $0$ so must be $2^{\frac{k}{k-1}}$. Third, it will alternate between being above $2^{\frac{k}{k-1}}$ one term and below $2^{\frac{k}{k-1}}$ the next, getting arbitrarily close to $2^{\frac{k}{k-1}}$.
Lemma 1: For some $n\in\mathbb{N}$, $a_n \geq 1$.
Proof: Suppose that this is not the case, so we have some least upper bound $B<1$ for the sequence $a_n,n\in\mathbb{N}$. Let $a_n$ be such that $a_n>B/2$. Clearly each term is positive and less than $1$ meaning $a_i^{1/k}>a_i$, so $a_{n+1}=a_n^{1/k}+a_{n-1}^{1/k}>a_n^{1/k}>a_n>B/2$ and thus $a_{n+2}=a_{n+1}^{1/k}+a_{n}^{1/k}>a_{n+1}+a_n>B/2+B/2=B$, contradicting $B$ being an upper bound.
Lemma 2: If $a_n > 2^{\frac{k}{k-1}}$, then $a_{n+1}< \max\{a_n,a_{n-1}\}$.
Proof: Note $x>2^{\frac{k}{k-1}}\implies x^{1/k}<\frac{x}{2}$, so if $a_n > 2^{\frac{k}{k-1}}$, $a_{n+1}< 2\max\{a_n,a_{n-1}\}^{1/k}<\max\{a_n,a_{n-1}\}$.
Lemma 3: If $a_0=a_1=1$, the sequence $a_n,n\in\mathbb{N}$ is increasing and has limit $2^{\frac{k}{k-1}}$.
Proof: We shall induct on $n$ from $2^{\frac{k}{k-1}}\geq a_2 = 2\geq a_1\geq a_0$. If $2^{\frac{k}{k-1}}\geq a_n\geq a_{n-1}\geq a_{n-2}$ then $a_{n+1}\geq 2a_n^{1/k}\geq 2\frac{a_n}{2}=a_n$, while $a_{n+1}\leq (2^{\frac{k}{k-1}})^{1/k}+(2^{\frac{k}{k-1}})^{1/k}=(2^{\frac{k}{k-1}})^{1/k}$, thus the sequence is increasing and bounded above, so must have a limit. This limit is either $0$ or $2^{\frac{k}{k-1}}$, and since $a_n\geq 1,\forall n\in \mathbb{N}$ it must be $2^{\frac{k}{k-1}}$
Proof of Theorem: By lemma 1, we have some $a_n>1$. We must also have $a_{n+1}>1$, so applying lemma 3 in light of the fact that $a_{n-1}^{1/k}+a_{n-2}^{1/k}$ is increasing in both $a_{n-1},a_{n-2}$ gives us that the limit inferior $I$ is at least $2^\frac{k}{k-1}$. Suppose we have some $a_{i-1},a_i>2^\frac{k}{k-1}$, so by lemma 2 $a_{i+1}< \max\{a_i,a_{i-1}\}$, but at the same time $a_{i+1}>(2^\frac{k}{k-1})^{1/k}+(2^\frac{k}{k-1})^{1/k}=2^\frac{k}{k-1}$ thus by induction the sequence $a_n,n\in\mathbb{N}$ is eventually decreasing and bounded below by $2^\frac{k}{k-1}$. Then it has a limit, and this limit must be $2^\frac{k}{k-1}$. If no such $a_{i-1},a_i$ exist, then we either have some $a_{i-1},a_i<2^\frac{k}{k-1}$ or all $a_{i-1},a_i$ are on different sides of $2^\frac{k}{k-1}$. In the first case, an argument analogous to the proof of lemma 3 shows that the sequence is has limit $2^\frac{k}{k-1}$. In the second, we can let $a_i>2^\frac{k}{k-1}$, so every term in the sequence $a_i,a_{i+2},\ldots$ is greater than $2^\frac{k}{k-1}$ while every term in $a_{i+1},a_{i+3},\ldots$ is less than $2^\frac{k}{k-1}$. By lemma 2 the first sequence is decreasing, so the limit superior $S$ of $a_n,n\in\mathbb{N}$ exists. It cannot be strictly greater than $2^\frac{k}{k-1}$, as if $S=2^\frac{k}{k-1}+\epsilon/2$ for $\epsilon>0$ we have some $a_{i+2j}<2^\frac{k}{k-1}+\epsilon$ so $a_{i+2j+2}=a_{i+2j}^{1/k}+a_{i+2j+1}^{1/k}<2^\frac{k}{k-1}/2+\epsilon/2+2^\frac{k}{k-1}/2 = S$, which contradicts the fact that the sequence $a_i,a_{i+2},\ldots$ is decreasing. Hence $I \geq 2^\frac{k}{k-1}\geq S$, so $I = S = 2^\frac{k}{k-1}$ thus $\lim\limits_{n \to \infty}{a_n}=2^{\frac{k}{k-1}}$.
