For integral K any Pisot-Vijayaraghavan number will do the trick (Qiaochu gives an example of one in the comments) - and there are a lot of these, although countably many. Regardless, these are good examples of how arbitrary non-integers can satisfy your condition.
A not-terribly-complicated proof of the fact that any PV number satisfies your condition for $K = 1$ is as follows. Consider a PV number $\alpha$ and its minimal polynomial p (in $\mathbb{R}/\mathbb{Q}$). Let the roots of p be $\alpha = x_1, x_2, \cdots, x_k$. By the fundamental theorem of symmetric polynomials, we know that the expression $P_n = x_1^n + x_2^n + \cdots + x_k^n$ is an integral linear combination of the symmetric polynomials in $x_1, x_2, \cdots, x_k$ - which, by Vieta's formulas, are simply the coefficients of p (as the minimal polynomial is monic). As p has integral coefficients, $P_n = x_1^n + x_2^n + \cdots + x_k^n$ is an integer for all n. As $\alpha$ is a PV-number, we know that $|x_2|, |x_3|, \cdots, |x_k| < 1$. So we have for large n that $|P_n - \alpha^n| < |x_2|^n + |x_3|^n + \cdots + |x_k|^n << 1,$ that is, $P_n - \alpha^n \rightarrow 0$ as desired.
This proof will work for any integer K - as for non-integer K, unless I'm missing something obvious (which could very well be the case) this is quite a bit harder. You might be able to use a nice trick to get rationals, but I suspect irrationals are a real pain.