$(C_{[0;1]}, \Vert . \Vert_1)$ is not a Banach space I'm going to prove that $C_{[0;1]}$ is not a Banach space w.r.s to the norm $\Vert x \Vert_1 = \int_{0}^{1} |x(t)| dt$ by consider the series $\sum_{n=1}^{\infty}x_n$ where $x_{n}(t)= t^{n} \cdot \sqrt{1-t}$. 

By MCT we shall easy check that this series is absolutely convergence. Indeed $$ \begin{array}{ll}\sum_{1}^{\infty}\Vert x_n \Vert_1 &amp; = \sum_{1}^{\infty}\int_{0}^{1}x_{n}(t)dt = \int_{0}^{1} \sum_{1}^{\infty} x_n(t)dt=\int_{0}^{1}\sum_{1}^{\infty}t^n\sqrt{1-t}\, dt \\ &amp;= \int_0^1 \frac{t}{\sqrt{1-t}}dt = \frac{4}{3}  \end{array} $$
But I'm stuck in proving this series is not convergence in $(C_{[0;1]}, \Vert . \Vert_1)$. I think that if this series is convergence in $(C_{[0;1]}, \Vert . \Vert_1)$ to $x$, it may convergence to $x$ pointwise, but I can't prove this.

Thank you for your help.
Notice that, if $g(t)=\sum_1^\infty x_n(t)$, then, for $t\in (0,1)$
$$
g(t)= \frac{\sqrt{1-t}}{1-t}.
$$
You can prove this as you did, using the fact that we have a geometric series. On the other hand, since $x_n(0)=0$ for all $n$, we have $g(0)=0$, which is impossible since the sum representation above forces $g(0)=1$. Therefore the series doesn't converge in your space.
Your series are convergence in this norm indeed, and you have already proven it. For an example to show the noncompleteness, consider fn(x) to be a function such that it equal to 1 when x &lt; 1/2, it equal to 0 when x > 1/2 + 1/n, and it is linear when 1/2 &lt; x &lt; 1/2 + 1/n. If such fn converges to a function f in the norm above, then view it as L1 we get a subsquence converges a.e. to f, thus f must be the function equal to 1 when x &lt; 1/2 and equal to 0 when x > 1/2 a.e., and cannot be continuous.