This is a follow up of [this question](https://mathoverflow.net/questions/178276/r-is-a-right-multiplier-and-rab-a-overset-implies-a-is-unital). Let $I$ be closed left ideal of $C^\*$-algebra $A$.


Assume we are given a sequence of left $A$-module morphisms $R\_n:I\to A$ with $\sum\_n \Vert R\_n\Vert<\infty$ and a sequence $\{b\_n\}\_{n\in\mathbb{N}}$ in the unit ball of $I$ with the property
$$
\sum\_n R\_n(a)b\_n=a\tag{1}
$$
for all $a\in I$. Is it true that $I=Ap$ for some $p\in I$. 


If the sum in $(1)$ would be finite the result would be true, since all finitely generated left ideals of $C^\*$-algebras are principal. Clearly, this is also true for $A$ being commutative, because we can set $p=\sum\_n R\_n(b\_n)$.



Is this left ideal of C*-algebra principal?
Yes, $I=Ap$ for some projection $p$. More generally, suppose that a closed left ideal 
$I$ contains a sequence $c\_1,c\_2,\dots$ such that 
$
\sum\_{i=1}^\infty \|c\_i\|<\infty
$
and for any $x\in I$ there exist $x\_1,x\_2,\dots$ such that $\|x\_i\|\leq \|x\|$ and 
$$
\sum\_{i=1}^\infty x\_ic\_i=x.
$$
Then $I$ is generated by a projection. (In the question, $c\_i=\|R\_i\|b\_i$ and 
$x\_i=R\_i(x)/\|R\_i\|$.) 


Here is a proof: Let $c=\sum\_{i=1}^\infty (c\_i^\*c\_i)^{\frac 1 2}$ (which converges). Since $(c\_i^\*c\_i)^{\frac 1 2}\leq c$, we have $c\_i=d\_ic$ for some contraction $d\_i\in A^{\*\*}$.
Choose $\epsilon>0$ small enough ($\epsilon=\frac 1 6$ will suffice). Choose $N$ such that 
$\sum\_{i>N}^\infty \|c\_i\|<\epsilon$. Since $c^{1/k}\in I$ for all $k=1,2,\dots$, we must have
$$
c^{\frac 1 k} =\sum\_{i=1}^N x\_ic\_i+\sum\_{i>N} x\_ic\_i=Xc^{\frac 1 2}+\Delta,
$$
where $X=\sum\_{i=1}^Nx\_id\_i$ and $\|\Delta\|\leq \sum\_{i>N} \|x\_i\|\|c\_i\| \leq \epsilon \|c\|^{1/k}$. Then,
\begin{align\*}
c^{\frac 2 k} &=c^{\frac 1 2}X^\*Xc^{\frac 1 2}+ \Delta'\leq \|X\|^2c+\|\Delta'\|\cdot 1.
\end{align\*}
The norms of $X$ and $\Delta'$ can be estimated: $\|X\|\leq \sum\_{i=1}^N\|x\_id\_i\|\leq N\|c\|^{1/k}$ and $\|\Delta'\|\leq 5\epsilon \|c\|^{2/k}$ (probably not a sharp estimate; I'm skipping the details). So
$$
c^{\frac 2 k}\leq N^2\|c\|^{\frac 2 k}\cdot c+ 5\epsilon \|c\|^{\frac 2 k}\cdot 1.
$$
This inequality must hold in $C^\*(c,1)$ for all $k$. For $k$ large enough, $\|c\|^{2/k}$ is close to 1, so the right hand side is bounded by a linear function in $c$ which at 0 is less than $6\epsilon$. So $0$ can only be an isolated point of the spectrum of $c$. Now,
with $p\in I$ the support projection of $c$, we have $c\_ip=d\_icp=d\_ic=c\_i$ for all $i$, and so $xp=x$ for all $x\in I$.


