Is it possible to swap sum and integration signs in this integral $$\int\limits_{0}^{+\infty}\sum\limits_{n=1}^{+\infty}a^nx^{1/2}e^{-nx}dx=\sum\limits_{n=1}^{+\infty}a^n\int\limits_{0}^{+\infty}x^{1/2}e^{-nx}dx$$ provided that $a∈(0;1]$? If it is possible, how to prove the uniform convergence of the series?
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See the limit theorems that you know for integrals. Does one (or more) of them apply in this case? – GEdgar Apr 15 '19 at 19:57
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I used the theorem of Weierstrass, $$\dfrac{a^nx^{1/2}}{e^{nx}}<\dfrac1{\sqrt{n}}$$ but sum $$\sum\limits_{n=1}^{+\infty}\dfrac1{\sqrt{n}}$$ diverges – Кирилл Колокольцев Apr 15 '19 at 20:30