I am studying the sequence of functions defined by $$u_n(x):=e^{-nx}-2e^{-2nx}$$ for $x>0$
I proved that $$\sum_{n\geq 1}\int_{]0,+\infty [}u_n(x)\mathrm{d}x= 0$$ and $$\int_{]0,+\infty[}\left(\sum_{n\geq 1}u_n(x)\right)\mathrm{d}x=\log(2).$$
However, I don't understand why can't I utilize the Beppo-Levi theorem to the sequence of functions defined by $$f_n(x)=\sum_{k=1}^n u_k(x).$$
I did a little code in Java to test the monotonicity of $f_n(x)$ for fixed values of $x$ and it really seems that for each $x>0$, $f_n(x)$ is a crescent sequence.
Which condition of Beppo-Levi fails here?
Suppose $|\sum_{k=1}^{n} u_k(x)| \leq g(x)$. Put $e^{-x}=t$, let $n \to \infty$ and compute the geometric sums $t+t^{2}+t^{3}+...$ and $t^{2}+t^{4}+t^{6}+...$ explicitly. You will get $g(x) \geq \frac {|e^{-x}-e^{-3x}-2e^{-2x}|} {1-e^{-2x}}=\frac {|e^{x}-e^{-x}-2|} {e^{2x}-1}$. As $x \to 0$ the numerator $\to 2$. Now using L\Hopital's Rule observe that $\frac {e^{2x}-1} x \to 2$ as $x \to 0$. Hence there exists $c>0$ and $\delta >0$ such that $g(x) > \frac c x$ for $0<x<\delta$. It follows that $\int_0^{\delta} g(x)dx =\infty$.