Equivalent condition to boundedness of a linear operator I have to prove the following statement:

Let $X$ and $Y$ be normed spaces over $\mathbb{K}$ and $T:X\to Y$ a linear operator. Prove that $T$ is bounded iff $\sum_{n=1}^\infty T(x_n)$ converges for all sequences $\{x_n\}$ for which $\sum_{n=1}^\infty x_n$ converges absolutely.

I have been stuck on this problem for days, yet I feel I'm really close to solving it. Any hint, idea or advice would be appreciated. I haven't been able to prove any of the implications, but this is what I have right now:
For the $(\implies)$ direction, assume that $T$ is bounded, and let $\{x_n\}$ be a sequence such that $\sum_{n=1}^\infty x_n$ converges absolutely. I'm trying to prove that $\sum_{n=1}^\infty T(x_n)$ converges on two steps. First I prove that the sequence $\{S_k\}_{k=1}^\infty$ where
$$
S_k=\sum_{n=1}^kT(x_n)
$$
is Cauchy. This is easy to show using the boundedness and linearity of $T$, and the fact that $\{\sum_{n=1}^k ||x_n||\}_{k=1}^\infty$ is Cauchy, because it converges. Then I would like to prove that the sequence $\{S_k\}$ has a convergent subsequence, as that would imply the result. However, I don't know how to show this. The only idea I had was to prove and use this statement:

If $\{x_n\}$ is a sequence on a normed space $X$ such that $\sum_{n=1}^\infty x_n$ converges absolutely, then there exists a sequence $\{n_k\}$ of positive integers such that the sequence $\{\sum_{n=1}^{n_k} x_n\}_{k=1}^\infty$ converges.

as the continuity and linearity of $T$ would then imply what I want. However, I don't think that statement is true, as it would then imply that every normed space is Banach, which is obviously not true. So I have no idea on how to proceed now. In general I am having trouble to prove the convergence of almost anything, any advice for that would be really helpful.
For the ($\impliedby$) direction, I am trying to use the sequential criterion for continuity to prove that $T$ is continuous, and since $T$ is linear that would prove that $T$ is bounded. Let $a\in X$ and $\{x_n\}$ a sequence in $X$ that converges to $a$. I want to prove that $T(x_n)\to T(a)$. I trying to use the same steps as in the other implication, i.e. I need to show that $\{T(x_n)\}$ is Cauchy and has a subsequence that converges to $T(a)$. Here however, I managed to show the existence of a convergent subsequence, but I don't see how to show that the sequence is Cauchy. To see that it has a subsequence convergent to $T(a)$, notice the following:
As $x_n\to a$, for each $k\in\mathbb{Z}^+$ there exists an $n_k\in\mathbb{Z}^+$ such that
$$
||x_{n_k}-a||<\dfrac{1}{2^k}
$$
And since $\sum_{k=1}^\infty 2^{-k}$ converges, we have that $\sum_{k=1}^\infty||x_{n_k}-a||$ converges. By hypothesis, this implies that $\sum_{k=1}^\infty T(x_{n_k}-a)$ converges, and from here we get that $||T(x_{n_k})-T(a)||=||T(x_{n_k}-a)||\to 0$ as $k\to\infty$, and as such $T(x_{n_k})\to T(a)$.
If now I manage to prove that $\{T(x_n)\}$ is Cauchy, I would get the result. However I have no idea on how to do this.
Thanks in advance for any help you could give me!
 A: If $T$ is bounded (and linear), and $\sum_{n=1}^\infty x_n$ converges to $x$ (no need for absolutely) then
$$T(x) = T\left(\sum_{n=1}^\infty x_n\right) = T\left(\lim_{N \to \infty} \sum_{n=1}^N x_n\right) = \lim_{N \to \infty} T\left(\sum_{n=1}^N x_n\right) = \lim_{N \to \infty} \sum_{n=1}^N T(x_n) = \sum_{n=1}^\infty T(x_n).$$
That's the "easy" direction ($\implies$).
As for ($\impliedby$), your partial result could actually be used to get you the full way, except that it tacitly assumes completeness. Just showing that $\sum \|x_{n_k} - a\|$ is finite doesn't actually prove $\sum (x_{n_k} - a)$ converges! Indeed, this implication is equivalent to the domain being complete.
The key observation is that, for linear operators, dealing with sequences is very similar to dealing with series. For a sequence $y_n \to y$, we can define $x_n = y_{n+1} - y_n$, and $\sum_{n=1}^\infty x_n = y - y_1$. This way, we can turn a convergent sequence into a convergent (telescoping) series.
The big wrinkle here is that $\sum_{n=1}^\infty x_n$ may not converge absolutely. To correct this, we can take a subsequence $y_{n_m}$ such that $\|y_{n_m} - y\| < \frac{1}{2^m}$. Then,
$$\|y_{n_{m+1}} - y_{n_m}\| \le \|y_{n_{m+1}} - y\| + \|y_{n_m} - y\| < \frac{1}{2^{m+1}} + \frac{1}{2^m} = \frac{3}{2^{m+1}}.$$
If we let $x_m = y_{n_{m+1}} - y_{n_m}$, then $\sum x_m$ converges absolutely to $y - y_{n_1}$. Using the condition on $T$, this means
$$T(y) -T(y_{n_1}) = \sum_{m=1}^\infty T(x_m) = \lim_{M \to \infty} \sum_{m=1}^M (T(y_{n_{m+1}}) - T(y_{n_m})) = \lim_{M \to \infty} (T(y_{n_{M+1}}) - T(y_{n_1})),$$
hence
$$T(y) = \lim_{M \to \infty}T(y_{n_{M+1}}).$$
This puts us in a similar position to where your solution took us. But, we can now use another auxiliary result to get us to the finish line:

Suppose $x_n$ is a sequence in a normed linear space (or indeed metric space). Further, suppose we have a fixed point $x$ and $x_n$ has the property that, given any subsequence $x_{n_m}$, there exists a (sub)subsequence $x_{n_{m_k}}$ converging to $x$. Then $x_n \to x$.

The proof is not too hard. If $x_n$ does not converge to $x$, then a subsequence must stay some $\varepsilon > 0$ distance away from $x$. Such a subsequence has no (sub)subsequences that converge to $x$.
Now, we can apply the above result to $T(y_n)$. Note that any subsequence $T(y_{n_m})$ corresponds to a subsequence $y_{n_m}$ of $y_n$. Such a subsequence still converges to $y$, so by the previous working, a (sub)subsequence $y_{n_{m_k}}$ exists such that
$$\lim_{k \to \infty} T(y_{n_{m_k}}) = T(y).$$
By the stated result, this implies $T(y_n) \to T(y)$ as well, hence $T$ is continuous (and bounded).
