Is it true that $\aleph(A_n)<c\implies \sum_{n\in\mathbb{N}}\aleph(A_n)<c$?

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Let $\{A_n\mid n\in \mathbb{N}\}$ be disjoint family of sets such that $\forall n\in\mathbb{N},\; \aleph(A_n)<c$. Is it true that $\aleph(\bigcup_{n\in\mathbb{N}}A_n)=\sum_{n\in\mathbb{N}}\aleph(A_n)<c$?

Here, $c=2^{\aleph_0}$.

I've been trying for a while a can't seem to be able to prove it. If it's true, a proof or a guide for a proof would be highly appriciated.