If for every $x=(x_n)\in l_p$ $(1\le p<\infty)$ the series $\sum a_n x_n$ converges then $a=(a_n) \in l_q$ where $\frac{1}{p}+\frac{1}{q}=1$.
For $1<p<\infty$ let $A_n(x)=\displaystyle\sum_{k=1}^{n} a_k x_k$ for all $x \in l_p$ and $n \in N$. Then $A_n$ is linear and, using Holder's Inequality, bounded on $l_p$. By hypothesis $(A_n(x))$ converges to, say, $A(x)$.
By the Banach-Steinhaus Theorem, as $l_p$ is a Banach space, $A\in l_p^{*}$, the dual of $l_p$.
Fix $r\in N$.
For $1\le k \le r$ let $x_k= $ sgn$ (a_k) |a_k|^{q-1}$ and for $k>r$ let $x_k=0$.
Then $x=(x_k) \in l_p$ with $||x||=\displaystyle \left( \sum_{k=1}^{r}|a_k|^{(q-1)p}\right)^\frac{1}{p}=\displaystyle \left( \sum_{k=1}^{r}|a_k|^{q}\right)^\frac{1}{p}$.
Here sgn$(z)=\frac{|z|}{z}$ for $z\neq 0$, sgn$(z)=1$ for $z=0$.
$|A(x)|=|\sum a_k x_k|=\displaystyle \sum_{k=1}^{r}|a_k|^{q}\le ||A|| ||x||$.
Hence either $\displaystyle \sum_{k=1}^{r}|a_k|^{q}=0$ or $\displaystyle \left(\sum_{k=1}^{r}|a_k|^{q}\right)^\frac{1}{q}\le ||A||$ which also follows if $\displaystyle \sum_{k=1}^{r}|a_k|^{q}=0$.
Letting $r \to \infty$, by the MCT, we have $\displaystyle \left(\sum_{k=1}^{\infty}|a_k|^{q}\right)^\frac{1}{q}\le ||A|| < \infty$ and so $a \in l_q$.
The case for $p=1$ is similar to the above but we use $x=e_n$, the nth unit vector, to extract the result $a \in l_{\infty}$.
Let $\{x_n\}$ be a Cauchy sequence. We need to show that it converges.
It suffices to show that $\{x_n\}$ possesses a converging subsequence.
As it is Cauchy, for every $\varepsilon>0$, and in particular for $\varepsilon=2^{-k}$, there exists an $N=N(k)$, such that, $m,n\ge N(k)$ implies that
$$
\|x_m-x_n\|<2^{-k}.
$$
The $N(k)$'s can be chosen to form a strictly increasing sequence.
Now we shall show that the subsequence $y_k=x_{N(k)}$ converges.
Note that $\|y_k-y_{k-1}\|<2^{{-k}}$.
In particular,
$$
y_k=y_1+(y_2-y_1)+(y_3-y_2)+\cdots+(y_k-y_{k-1})=z_1+z_2+\cdots+z_k,
$$
and for $z_k$ we have that
$$
\sum \|z_k\|\le \|y_1\|+\sum_{k=1}^\infty 2^{-k}=\|y_1\|+\frac{1}{2}<\infty,
$$
and thus $y_k$ converges. Let $y_k\to y$.
It is left to you to show that $x_n$ converges as well to $y$.
Best Answer
Assume $2.$
Take a convergent series $\displaystyle\sum x_n.$ Let $\rho_0=0$ and $\displaystyle r_n=\sum_{k=n}^\infty x_k.$ Then $r_n\to 0.$ We have \begin{multline*}\sum_{k=1}^n \rho_k(x_k)=\sum_{k=1}^n \rho_k(r_{k}-r_{k+1})= \sum_{k=1}^n \rho_k(r_k)-\sum_{k=1}^n \rho_k(r_{k+1})\\ = \sum_{k=1}^n \rho_k(r_k)-\sum_{k=2}^{n+1}\rho_{k-1}(r_k) = \sum_{k=1}^{n}[\rho_k-\rho_{k-1}](r_k)-\rho_n(r_{n+1}) \end{multline*} By assumptions we have $$\|\rho_n\|\le \sum_{k=n}^\infty \|\rho_k-\rho_{k+1} \|\le \sum_{k=1}^\infty \|\rho_k-\rho_{k+1} \|=:C $$ Hence $|\rho_n(r_{n+1})|\le C\|r_{n+1}\|\to 0.$ The series $$\sum_{k=1}^\infty [\rho_k-\rho_{k-1}](r_k)$$ is absolutely convergent as $$\sum_{k=1}^\infty |[\rho_k-\rho_{k-1}](r_k)|\le \sum_{k=1}^\infty \|\rho_k-\rho_{k-1}\|\|(r_k)\| $$ and $\|r_k\|$ is a bounded sequence. Therefore the original series is absolutely convergent and $$\sum_{k=1}^\infty \rho_k(x_k)=\sum_{k=1}^\infty [\rho_k-\rho_{k-1}](r_k)$$
Assume $1.$
Then $$\sup_n\|\rho_n\|<\infty$$ Indeed, assume for a contradiction that $$\sup_n\|\rho_n\|=\infty$$ Then there exists a sequence of nonnegative numbers $a_n$ such that $$\sum_{n=1}^\infty a_n<\infty, \qquad \sum_{n=1}^\infty a_n\|\rho_n\|=\infty$$ For every $n$ there exists $u_n$ such that $$ \|u_n\|=1,\quad \rho_n(u_n)\ge {1\over 2}\|\rho_n\|$$ Let $x_n=a_nu_n.$ Then the series $\sum x_n$ is convergent $$\sum_{n=1}^\infty \|x_n\|=\sum_{n=1}^\infty a_n<\infty, \quad \sum_{n=1}^\infty \rho_n(x_n)= \sum_{n=1}^\infty a_n\rho_n(u_n)\ge {1\over 2}\sum_{n=1}^\infty a_n\|\rho_n\|=\infty$$ which gives a contradiction with assumption $1.$
Assume $2.$ does not hold. Then there exists a sequence $b_n\to 0^+$ such that $$\sum_{k=1}^\infty b_k\,\|\rho_{k+1}-\rho_{k}\|=\infty$$ There exist elements $y_k$ such that $$\|y_k\|=1,\qquad \rho_{k+1}(y_k)-\rho_{k}(y_k)\ge {1\over 2}\|\rho_{k+1}-\rho_{k}\|.$$ Hence $$\sum_{k=1}^\infty b_k\,[\rho_k(y_{k+1})-\rho_{k}(y_k)]=\infty\qquad (*)$$ Let $y_0=0,$ $b_0=0$ and $x_n=b_{n-1}y_{n-1}-b_{n}y_{n}.$ As $b_ny_n\to 0$ the series $\displaystyle\sum_nx_n$ is convergent (to $0$). We have \begin{multline*}\sum_{k=1}^n \rho_k(x_k)=\sum_{k=1}^n b_{k-1}\rho_{k}(y_{k-1})-\sum_{k=1}^n b_{k}\rho_k(y_{k})\\ =\sum_{k=1}^{n-1}b_k\rho_{k+1}(y_k)-\sum_{k=1}^{n}b_k\rho_{k}(y_k)=\sum_{k=1}^n b_k [\rho_{k+1}(y_k)-\rho_k(y_k)]- b_n\rho_{n+1}(y_n) \end{multline*} Since the norms of $\rho_n$ are uniformly bounded, $b_n\to 0$ and $\|y_n\|=1,$ we get $b_n\rho_{n+1}(y_n)\to 0.$ By $(*)$ the series $$\sum_{n=1}^\infty \rho_n(x_n)$$ is divergent, which gives a contradiction.