# Conditional Schauder basis

Let's consider $$\ell_1$$ space with the standard unconditional Schauder basis $$(e_n)_{n=1}^\infty$$ and let's define the following sequence: $$x_1:=e_1$$ and $$x_n := e_{n-1} - e_n$$ for $$n >1$$. I found this sequence also mentioned on this site, as being an example of a conditional Schauder basis for the space $$\ell_1$$. So this is what I wanted to check. Let's fix $$x \in \ell_1$$. Then we can write $$x=\sum_{n=1}^\infty a_n e_n$$ since $$(e_n)_{n=1}^\infty$$ is a basis. Then, by doing some series manipulation and working on $$x_n$$ we can define the following sequence: $$b_1:=0$$ and $$b_n:=\sum_{k=1}^{n-1} a_k.$$ Then we have that $$x=\sum_{n=1}^\infty b_n x_n$$ so this shows that $$(x_n)_{n=1}^\infty$$ is a Schauder basis for $$\ell^1$$ (hopefully here I didn't make any mistakes). Now, in order to show that it's conditional I wanted to use one of the characterizations, namely either find a sequence $$(\varepsilon_n)_{n=1}^\infty \subset \{-1,1\}$$ such that $$\sum_{n=1}^\infty \varepsilon_n x_n$$ doesn't converge in $$\ell_1$$, or equivalently $$(\lambda_n)_{n=1}^\infty \subset \mathbb{K}$$ such that $$\sup_n |\lambda_n| < \infty$$ and again, such that $$\sum_{n=1}^\infty \lambda_n x_n$$ doesn't converge in $$\ell_1$$ but I'm not sure what kind of a sequence we can choose here. Could anyone share some insight how we can choose such a sequence?

Thank you!

Just take $$\epsilon_n=(-1)^{n}$$ and compute the partial sums of $$\sum \epsilon_n x_n$$. You wil see easily that the series does not converge.