For n=1,2,... let $f_n: \mathbb{R} \to \mathbb{R}$ defined by
$$f_n(x)= {\sqrt[3]{n-x} \over {e}^{(x-n)^2}}$$
Determine to what intervals of the real axis $I$ the series $\sum_{n=1}^\infty f_n$ converges in norm $L^1 (I)$
My work:
$\int_a^b |\sum_{n=1}^{\infty}f_n(x)-\sum_{n=k-1}^{\infty}f_n(x)|dx=\int_a^b |\sum_{n=k}^{\infty}f_n(x)| dx \le \int_a^b \sum_{n=k}^{\infty}{\sqrt[3]{|n-x|} \over {e}^{(x-n)^2}} dx $ $\le \sum_{n=k}^{\infty} \int_{a-n}^{b-n} {\sqrt[3]{|y|} \over {e}^{y^2}} dy=\sum_{n=k}^{\infty}a_n$
if $b \neq +\infty$ :$a_n \le \int_{a-n}^{b-n} {-y \over e^{y^2} }dy= - {1\over2}[-e^{-(b-n)^2}+e^{-(a-n)^2}]$
Therefore $\sum_{n=k}^{\infty}a_n$ coverges and $\sum_{n=1}^\infty f_n$ converges in norm $L^1 ((-\infty,b))$ $ \forall b \in \mathbb{R}$
I think that if $b=+\infty$ and $a \in \mathbb{R} \cup \{-\infty\} $ the series diverges but I don't know how to show it.