If $T$ is a bounded linear map and $\sum x_n$ is an absolutely convergent series, then $T(\sum x_n) = \sum T(x_n)$ Is the following true? If so, how to prove it?

If $T:X \to Y$ is a bounded linear map between the Banach space $X$ and the normed vector space $Y$ and $\sum x_n$ is an absolutely convergent series, then $T(\sum x_n) = \sum T(x_n)$.

I need this claim to finish a proof that if $X$ is a Banach space and $M$ is a proper closed subspace, them $X/M$ is a Banach space, but I am not able to show it.
Thanks in advance and kind regards.
On a Banach space $X$ a linear operator $F$ is continuous if and only if it's bounded.
For your case, if it is bounded you have that $\exists L$ such that for any $x\in X$ $\Vert F x\Vert \leq L \Vert x \Vert$.
Hence you have that if $x_n\to x$, fixed any $\epsilon$, for $n$ large enough $\Vert x_n-x\Vert\leq \epsilon$.
Then $\Vert Fx_n - F x\Vert = \Vert F(x-x_n)\Vert \leq L\Vert x-x_n\Vert \leq L\epsilon$.
Now an absolutely convergent series is a converging sequence (of the partial sums) so your statement follows.
You can find detailed proofs of both facts as Lemma 4.4 and Theorem 4.5 in:
http://www.pitt.edu/~hajlasz/Notatki/Functional%20Analysis2.pdf