If $u_n $ converges to 9, then $\exists w_n$ that converges to $0$ such that $\sum u_n = \sum w_n $

Question

Suppose $D = \left \{ 0,1,2,3,4,5,6,7,8,9 \right \}$

Suppose $(u_n)_{n \in \mathbb{N}} $ such that $u_n : \mathbb{N} \rightarrow D$

Now I am asked to show that $v_n=\frac{u_n}{10^n}$ converges to a real number $x \in [0,10]$. As $|u_n| \leq 9$, then $|v_n| \leq \frac{9}{10^n}$, and $lim \frac{9}{10^n} = 0 $ when $n$ tends to infinity. Thus $u_n$ converges to a real $x \in [0,10]$.

Now I am asked to show that if $u_n$ converges to $9$, then $\exists w_n$ such that $w_n: \mathbb{N} \rightarrow D$, $w_n$ converges to 0 and that $\sum u_n = \sum w_n$.

This is my attempt:

If $u_n$ converges to 9, then $\exists N$, such that $\forall n \geq N$, $u_n=9$. Thus we have for $n \geq N$:

$\sum_{k=1}^{n}u_k=\sum_{k+1}^{N}u_k + 9(n-N)$

Also, as $w_n$ converges to $0$, then $\exists N'$, such that $\forall n \geq N'$, $w_n=0$. Thus we have: $\sum_{k=1}^{n}w_k=\sum_{k+1}^{N}u_w$

And so:

$$\sum_{k+1}^{N}u_k + 9(n-N) = \sum_{k+1}^{N'}w_k$$

And this is where I am stuck.