Let $(x\_{n})\_{n=1}^\infty$ be a bounded sequence in a Banach space $X$. We set
$$\textrm{ca}((x\_{n})\_{n=1}^\infty)=\inf\_{n}\sup\_{k,l\geq n}\|x\_{k}-x\_{l}\|.$$
Then $(x\_{n})\_{n=1}^\infty$ is norm-Cauchy if and only if $\textrm{ca}((x\_{n})\_{n=1}^\infty)=0$.


Let $X$ be a Banach space with a basis $(x\_{n})\_{n=1}^\infty$. Recall that $(x\_{n})\_{n=1}^\infty$ is boundedly complete if for every scalar sequence $(a\_{n})\_{n=1}^{\infty}$ with $\sup\_{n}\|\sum\_{i=1}^{n}a\_{i}x\_{i}\|<\infty$, the series $\sum\_{n=1}^{\infty}a\_{n}x\_{n}$ converges.


Let $(x\_{n})\_{n=1}^\infty$ be a basis for a Banach space $X$. We set
$$\textrm{bc}((x\_{n})\_{n=1}^\infty)=\sup\Big\{\textrm{ca}((\sum\_{i=1}^{n}a\_{i}x\_{i})\_{n=1}^\infty)\colon \|\sum\_{i=1}^{n}a\_{i}x\_{i}\|\leq 1, \forall n\Big\}.$$
Clearly, $(x\_{n})\_{n=1}^\infty$ is boundedly complete if and only if $\textrm{bc}((x\_{n})\_{n=1}^\infty)=0$.


We consider some classical non-boundedly complete bases as follows:


1. $\operatorname{bc}((e\_{n})\_{n})=1$, where $(e\_{n})\_{n}$ is the unit vector basis of $c\_{0}$.
2. $\operatorname{bc}((s\_{n})\_{n})=1$, where $(s\_{n})\_{n}$ is the summing basis of $c\_{0}$.
3. $\operatorname{bc}((e\_{n})\_{n=0}^{\infty})=2$, where $(e\_{n})\_{n=0}^{\infty}$ is the unit vector basis of $c$ ($e\_{0}=(1,1,1,\ldots)$).
4. $\operatorname{bc}((e\_{n})\_{n})=1$, where $(e\_{n})\_{n}$ is the unit vector basis of the James space $\mathcal{J}$.


Question 1. $\textrm{bc}((x\_{n})\_{n=1}^\infty)\in \{0,1,2\}$ for every basis $(x\_{n})\_{n}$ ?


Since $c\_{0}, C[0,1]$ or $L\_{1}[0,1]$ has no boundedly complete basis, we have the following natural questions:


Question 2. $\textrm{bc}((x\_{n})\_{n=0}^\infty)=$ ?, where $(x\_{n})\_{n=0}^\infty$ is the Faber-Schauder basis for $C[0,1]$ (See I. Singer, Bases in Banach spaces I, pp. 11 for the Faber-Schauder basis).


Question 3. $\textrm{bc}((h\_{n})\_{n=1}^\infty)=$ ?, where $(h\_{n})\_{n=1}^\infty$ is the Haar basis for $L\_{1}[0,1]$.


PS: We have proved that if $(x\_{n})\_{n}$ is an $1$-unconditional basis, then $\textrm{bc}((x\_{n})\_{n=1}^\infty)=0$ or $1$.



Quantifications of boundedly complete bases
This is an answer to Dongyang's query in the comments. I show $\text{bc}(x\_n)\ge 1$ if $(x\_n)$ is a monotone basis for $X$ that is not boundedly complete. Let $(x\_n^\*)$ be the functionals biorthogonal to $(x\_n)$ and let $Y$ be the closed linear span in $X^\*$ of $(x\_n^\*)$, so that $(x\_n^\*)$ is a monotone basis for $Y$. By monotonicity of $(x\_n)$, the natural evaluation map from $X$ into $Y^\*$ is an isometric isomorphism and, since $(x\_n)$ is not boundedly complete, this map is not surjective. Therefore we can regard $X$ as being a proper subspace of $Y^\*$. Thus there is a unit vector $y^\*$ in $Y^\*$ whose distance to $X$ is arbitrarily close to one (say, larger than $1-\epsilon$).
You can write $y^\*=\sum a\_n x\_n$ with the understanding that the convergence of the sum is in the weak$^\*$ topology on $Y^\*$. By monotonicity, for every $N$,
$\|\sum\_{n=1}^N a\_n x\_n\| \le 1 = \|y^\*\|$. On the other hand, for every $N$, $\|y^\* - \sum\_{n=1}^N a\_n x\_n\|$ is at least as large as the distance from $y^\*$ to $X$, which implies that for every $N$, $\|\sum\_{n=N}^\infty a\_n x\_n\| > 1-\epsilon$. Now for any fixed $N$, $\sum\_{n=N}^M a\_n x\_n$ converges weak$^\*$ to $\sum\_{n=N}^\infty a\_n x\_n$ and hence for large $M$,
$\|\sum\_{n=N}^M a\_n x\_n\| > 1-\epsilon$. From this it follows that
$\text{bc}(x\_n)\ge 1$.


