Sum of finite ordinals: $\lambda_1+\lambda_2+\dots+\lambda_n+\dots=\omega$ Prove: $\lambda_1+\lambda_2+\dots+\lambda_n+\dots=\omega$, where $\lambda_i$ are finite ordinal nonzero numbers.

I tried like this.

$A_i$ set and ord($A_i$)$=\lambda_i, i=\{1,2,...\}$ and ord($\mathbb{N}$)$=\omega$.

$\lambda_1+\lambda_2+\dots+\lambda_n+\dots=$ord($\bigcup (A_i$x$\{i\})$)

$\bigcup (A_i$x$\{i\})$ is countable union of finite set. And that is countable. 
So ord($\bigcup (A_i$x$\{i\})$)=ord($\mathbb{N}$)

Is this correct?

EDIT:
Definition: $\{\lambda_i\}_{i\in I}$ is family of ordinal numbers and $\lambda_i=$ord($A_i$). Then $\sum_{i\in I}=$ord($\bigcup_{i\in I} (A_i$x$\{i\})$)

I think that I need to find order isomorphism from $\bigcup_{i\in I} (A_i$x$\{i\})$ to $\mathbb{N}$ but I'm not sure how.

Given two posets $(S,\le_S)$ and $(T,\le_T)$, and order isomorphism from $(S,\le_S)$ to $(T,\le_T)$ is bijective function $f$ from $S$ to $T$ with the property that, for every $x,y\in S$, $x\le_S y$ iff $f(x)\le_T f(y)$
It is easy to show the sum is not finite. The sum is a limit of finite sums however. the only candidate is $\omega$.
Note $\lambda_{i} = 1 + 1 + 1 \cdots$ ($\lambda_{i}$ times). Replace $\lambda_{i}$ accordingly in the sum. By the associative property we can remove parentheses and get $(1 + 1 + 1 + \cdots)$ which is obviously $\omega$.