We have $b_{n}\leq\sqrt[n]{n}k$. Therefore, there exists $N\in\mathbb{N}$ such that $b_{n}\leq k$ when $n\geq N$. There exists $N_{1}>N$ such that $k^{n}>n\left(k-1\right)^{n}$ when $n>N_{1}$. Then we have $m^{n}>n\left(m-1\right)^{n}$ if $b_{n}=m\leq k$. This implies that there is at most one $a_{i}\ne 0$ in $a_{1}, a_{2}, \cdots, a_{n}$ when $b_{n}\in\mathbb{Z}$, $n\ge N_{1}$. If $b_{1}, b_{2}, \cdots$ have infinitely many integers, there is at most one non-zero element in $a_{1}, a_{2}, \cdots$.