Sum of bounded operators converge in B(X,Y) Let $X$ be a normed space and $Y$ a Banach space and let $(A_n)_n$ be a sequence of bounded operators from $X$  to $Y$, and let there exist a sequence of positive nombers $(c_n)_n$ so that $||A_n||\leq c_n$, for all $n \in \mathbb{N}$ and $\sum_{n=1}^{\infty}c_n<\infty$.
Prove that  $\sum_{n=1}^{\infty}A_n$ converges in $B(X,Y)$, space of all bounded operators from $X$ to $Y$.
I tried the following: Let $S_n$ be a partial sum of $\sum_{n=1}^{\infty}A_n.$ As $X$ is normed space, $Y$ Banach, $B(X,Y)$ is Banach space. So if I show that that the partial sum is Cauchy it will converge in $B(X,Y).$ 
For $m,n\in \mathbb{N}$, $n\leq m$, $x \in X$,
$||S_m(x)-S_n(x)||=||\sum_{k=n+1}^{m}A_n(x)||\leq \sum_{k=n+1}^{m}||A_n(x)||\leq \sum_{k=n+1}^{m}c_n=c_m-c_n<\infty$.
$\sum_{n=1}^{\infty}A_n(x)$ is bounded: $||\sum_{n=1}^{\infty}A_n(x)||\leq \sum_{n=1}^{\infty}||A_n(x)||$ and as every $A_n$ is bounded $||A_n(x)||\leq M ||x||, M>0$ we can put $M=c_n$, we have $\sum_{n=1}^{\infty}||A_n(x)||\leq\sum_{n=1}^{\infty}c_n||x||.$
Thank you for any kind of help.
 A: Since $\|A_n\|\leq c_n$ and $\sum_\limits{n=1}^{\infty}c_n <\infty$, it follows that $\sum_\limits{n=1}^{\infty}\|A\|_n <\infty$ and since $B(X,Y)$ is a Banach space, so, now your conclusion will follow from the following
$\textbf{Lemma}:$In a normed linear space $X$ absolute convergence implies convergence if and only if $X$ is a Banach space.
$\textit{Proof}:$ Let $\sum_{k = 1}^{\infty}x_k$ be a series in $X$. First let us assume that $X$ is a Banach space and $\sum_{k = 1}^{\infty}\|x_k\|$ converges. We need to show that $\sum_{k = 1}^{\infty}x_k$ is convergent. Let $(s_n)$ be the sequence of partial sums. Then $(s_n)$ is a Cauchy sequence since
\begin{equation}
\|s_{n+m} - s_n\| = \|\sum_{k = n+1}^{n+m}x_k\| \leq \sum_{k = n+1}^{n+m}\|x_k\|.
\end{equation}
Thus, $\sum_{k = 1}^{\infty}x_k$ is convergent as $X$ is a Banach space.
Conversely, let us assume that absolute convergence implies convergence. Let $(z_n)$ be a Cauchy sequence in $X$. Then there exists a subsequence $(z_{n_k})$ of $(z_n)$ such that
\begin{align*}
\|z_{n_{k+1}} - z_{n_k}\| < \dfrac{1}{2^k} & & (k \geq 1).
\end{align*}
Let $x_k = z_{n_{k+1}} - z_{n_k}, k \geq 1$. Then $\sum_{k = 1}^{\infty}\|x_k\|$ is convergent. Hence by hypothesis it follows that $\sum_{k = 1}^{\infty}x_k$ is convergent, i.e., $(z_{n_{k}})$ is convergent. Since, $(z_n)$ is a Cauchy sequence which has a convergent subsequence, it follows that $(z_n)$ is convergent.
