Let $f_n(x) = ae^{-nax}-be^{-nbx}$ where $0<a<b$.
Prove that $$ \sum_1^\infty f_n \in L^1([0,\infty),m)\quad\text{and}\quad\int_0^\infty\sum_1^\infty f_n(x)\,dx=\log(b/a). $$
This is the problem 27 of chapter 2 of Folland Real Analysis. Could anyone give me some helps about how to attack this problem? I can't think of any way...
Hint : $$\sum_{n=1}^{\infty}f_n(x)=\sum_{n=1}^{\infty}ae^{-nax}-\sum_{n=1}^{\infty}be^{-nbx}=a(\frac{e^{-ax}}{1-e^{-ax}})-b(\frac{e^{-bx}}{1-e^{-bx}})\in L^1([0, \infty),m)$$
Consider part (2) first, then for every $m > 0$, $ae^{-anx},be^{-bnx}$ are bounded on $[0,m]$ and we have $$ \int_{0}^{m} ae^{-anx} = \frac{1}{n}(1 - e^{-nam}) $$ for any $n$, by monotone convergence: $$ \int_{0}^{\infty} f = \lim_{m} \frac{1}{n} (e^{-nbm} - e^{-nam}) = 0 $$ and hence $\sum^{\infty} \int_{0}^{\infty} f = 0$.
Then consider part (1). For any $n$ and $x \geq \chi := \frac{\log (b/a)}{(b-a)n} > 0$, $ae^{-anx} \geq be^{-bnx}$ so $$ \int_{[0,\infty]} f \geq \int_{[\chi,\infty]} (ae^{-anx} - be^{-bnx}) = \frac{1}{n} (e^{-na\chi} - e^{-nb\chi}) = \frac{1}{n} C $$ for a constant $C = \frac{b}{a}(e^{-\frac{a}{b-a}} - e^{-\frac{b}{b-a}}) > 0$. Because the series $\sum_{n=1}^{\infty} \frac{1}{n} C$ diverges, by comparison, $\sum_{n=1}^{\infty} \int f$ diverges as well.
Then consider part (3). $\sum_{n=1}^{\infty} f_{n}(x) = \frac{a e^{-ax}}{1 - e^{-ax}} - \frac{b e^{-bx}}{1 - e^{-bx}}$, and $$ \vert{\sum_{n=1}^{\infty} f_{n}}\vert = \sum_{n=1}^{\infty} f_{n} $$because $$ \frac{a e^{-ax}}{1 - e^{-ax}} - \frac{b e^{-bx}}{1 - e^{-bx}} = \frac{e^{-bx}e^{-ax}}{(1-e^{-bx})(1-e^{-ax})}(ae^{bx} - be^{ax} - (a -b)) > 0 $$ by Taylor expansion of exponentials. Then $$ \int \vert{\sum_{n=1}^{\infty} f_{n}}\vert = \int \sum_{n=1}^{\infty} f_{n} = \lim_{m\to \infty} \log(\frac{1-e^{-b/m}}{1 - e^{-a/m}}) = \log \frac{b}{a} $$
