Challenging probability problem

Let $t_n = 1-\frac{1}{3^{3^{2n}}}$ and $K_n = 3^{3^{2n-1}}$ for $n \ge 1$. We claim that, for all $n$ large (maybe $n \ge 10$ is fine), with probability at least $1-\frac{1}{n^2}$, the sign of $\sum_{k \ge 1} u_kt_n^k$ is the same as the sign as $\sum_{K_n \le k < K_{n+1}} u_kt_n^k$. Note that $|\sum_{k \ge K_{n+1}} u_k t_n^k| \le \sum_{k \ge K_{n+1}} t_n^k \le \frac{t_n^{K_{n+1}}}{1-t_n} = (1-\frac{1}{3^{3^{2n}}})^{3^{3^{2n+1}}}3^{3^{2n}} \approx e^{-3^{3^{2n+1}}/3^{3^{2n}}}3^{3^{2n}} = e^{-3^{3^{2n}}}3^{3^{2n}} \approx e^{-3^{3^{2n}}}$. And, $|\sum_{k < K_n} u_kt_n^k| \le K_n = 3^{3^{2n-1}}$. Therefore, if $|\sum_{K_n \le k < K_{n+1}} u_kt_n^k| \ge 2\cdot 3^{3^{2n-1}}$ (say), then $\sum_{k \ge 1} u_k t_n^k$ has the same sign as $\sum_{K_n \le k < K_{n+1}} u_k t_n^k$. And with probability at least $1-\frac{1}{n^2}$, we'll have $|\sum_{K_n \le k < K_{n+1}} u_kt_n^k| \ge 2\cdot 3^{3^{2n-1}}$. Indeed, $\mathbb{E}\left[|\sum_{K_n \le k < K_{n+1}} u_k t_n^k|^2\right] = \sum_{K_n \le k < K_{n+1}} t_n^{2k} \ge 3^{3^{2n}}$, and we have concentration since everything is i.i.d.. So by Borel-Cantelli, with probability $1$, for all but finitely many $n$ we will have that $\sum_{K_n \le k < K_{n+1}} u_kt_n^k$ has the same sign as $\sum_{k \ge 1} u_kt_n^k$; clearly the signs of $\sum_{K_n \le k < K_{n+1}} u_kt_n^k$ are independent as $n$ ranges and each half $50-50$ probability of being positive or negative, so with probability $1$ we get infinitely many sign changes and thus infinitely many roots.