Norm convergence of series in banach space Let $\mathfrak{X}$ be a Banach space and $\left(\rho_{n}\right)_{n \in \mathbb{N}}$ be a sequence in $\mathfrak{X}^\#$. Prove that the following assertions are equivalent:

*

*For each norm-convergent series $\sum_{n=1}^{\infty} x_{n}$ in $\mathfrak{X}$, the scalar series $\sum_{n=1}^{\infty} \rho_{n}\left(x_{n}\right)$ converges;

*$\sum_{n=1}^{\infty}\left\|\rho_{n}-\rho_{n+1}\right\|<\infty$.

I was given the hint that I should use summation by parts for this. But I am unable to progess in either direction.
Assuming 2. let $s_n=\sum x_n$ be a convergent sequence to say $s$. WLOG $s=0$ then $s_n$ are bounded by say $M$ then we want to show that $\sum \rho_n(x_n)$ converges. We have
$$\sum |\rho_n(x_n)|\le \sum||\rho_n||.||x_n||\le M\sum||\rho_n||$$ So if I can get the boundedness of $\sum||\rho_n||$ from (2) then I can conclude (1). Maybe Banach-Steinhaus can help but I can't figure it out.
(1) implies (2) I have no ideas.
 A: 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.
