# Exchanging series and integrals

I know that I can use Lebesgue or monotone convergence theorem to exchange limit of partial sums and a Lebesgue integral, given a power series or a generic function series. But in general given a series $$\sum_{n=0}^{\infty}a_n$$ which converges, and defined $$\int_0^\infty\sum_{n=0}^{\infty}a_n f_n(u)du$$ with $$f_n(u)$$ integrable, I was wondering when I could exchange the integration and the series. In particular in the context of Borel summation , given $$\int_0^\infty e^{-u} \sum_{n=0}^{\infty}\frac{a_nu^n}{n!}du$$, I was wondering how could I demonstrate that if $$\sum_{n=0}^{\infty}a_n$$ converges, then I can exchange the integral and the series. (I know that for power series $$\sum_{n=0}^{\infty}a_n z^n$$ the work can be done using the radius of convergence and I can always find a dominant)

As suggested by Gerald Edgar, we can use the Fubini--Tonelli theorem. By the Tonelli theorem, $$\int_0^\infty \sum_{n=0}^{\infty}\Big|e^{-u} \frac{a_nu^n}{n!}\Big|\,du =\sum_{n=0}^{\infty}\frac{|a_n|}{n!}\int_0^\infty e^{-u} u^n\,du =\sum_{n=0}^{\infty}|a_n|<\infty.$$ So, the Fubini theorem is applicable, that is, one can interchange the integral and the series.
• Being pedantic technically we should write the series as an integral over an appropriate measure and moreover we should demonstrate that $$\int_0^\infty \sum_{n=0}^{\infty}\Big| e^{-u}\frac{a_nu^n}{n!}\Big|\,du < \infty.$$ and because of integrability, we can swap the oringinal integral and sum. Now here $\int_0^\infty e^{-u} \sum_{n=0}^{\infty}\frac{|a_n|u^n}{n!}\,du =\sum_{n=0}^{\infty}\frac{|a_n|}{n!}\int_0^\infty e^{-u} u^n\,du$ are we using Tonelli's Thm. for non-negative measurable functions? But if the series is uniformly convergent can we use the Dominated Convergence Theorem? Oct 18, 2020 at 12:56