Find limit of $(a_n)$ if $a_n=\frac{1}{k}\sum\limits_{i=1}^{k}a_{n-i}$, $a_0=1$ and $a_{-n}=0$ for $n>0$

Your equation can be rewritten in the form of a renewal equation: $$ a_n = b_n +(a*p)_n $$ where, for every $n\ge0$, $$p_n =\frac{1}{k}1_{1\le n\le k}\qquad b_n=1_{n=0}$$ and $$ (a*p)_n :=\sum_{j=0}^n a_{n-j}p_j $$ Here, $\{b_j\}$ is added since $a_n =\frac{1}{k}\sum\limits_{j=1}^k a_{n-j}$ does not hold for $n=0$.

Now, it is a result of the (discrete version of) Blackwell's renewal theorem that $$ \lim_{n\to\infty} a_n =\frac{\sum\limits_{n=0}^\infty b_n}{\sum\limits_{n=0}^\infty np_n}=\frac{1}{\frac{1}{k}\frac{k(k+1)}{2}}=\frac{2}{k+1} $$