# Does $\int_1^\infty\frac{f_ng_n}{f_n^2+g_n}dx$ go to $0$ under these conditions of $f_n$ and $g_n$

Question: Let $$f_n,g_n:[1,\infty)\rightarrow (0,\infty)$$ be two sequences of measurable functions such that $$|g_n(x)|\leq\frac{1}{x^3} \forall x\geq1$$ and $$f_n\rightarrow 0$$ pointwise almost everywhere. Is it always true that $$\int_1^\infty\frac{f_ng_n}{f_n^2+g_n}dx\rightarrow0$$?

My thoughts: I was thinking that some sort of dominated convergence theorem will need to be used here since $$|g_n(x)|$$ is bounded by an integrable function $$\frac{1}{x^3} \forall x\geq1$$. The pointwise a.e. convergence of $$f_n$$ to $$0$$ seems like it would be easy to play with, but I'm just not sure how to deal with these functions as they are in the integrand. Any help, suggestions, etc. are appreciated! Thank you!

Note first that $$2f_n\sqrt{g_n}\leq f_n^2+g_n$$, therefore $$\frac{f_ng_n}{f_n^2+g_n}\leq\frac{\sqrt{g_n}}{2}\leq\frac{1}{2x^{3/2}}.$$ Hence, since $$\frac{1}{2x^{3/2}}$$ is integrable, for $$\varepsilon>0$$, there exists $$M_{\varepsilon}>0$$ such that $$\int_{M_{\varepsilon}}^{\infty}\frac{f_ng_n}{f_n^2+g_n}<\varepsilon.$$ Note now that, if $$f_n>1$$, then $$\frac{f_ng_n}{f_n^2+g_n}, while if $$f_n\leq 1$$, then $$\frac{f_ng_n}{f_n^2+g_n}\leq\frac{f_ng_n}{g_n}=f_n$$. Therefore, $$\int_1^{\infty}\frac{f_ng_n}{f_n^2+g_n}\leq\int_1^{M_{\varepsilon}}\frac{f_ng_n}{f_n^2+g_n}+\varepsilon\leq\int_{[1,M_{\varepsilon}]\cap[f_n>1]}g_n+\int_{[1,M_{\varepsilon}]\cap[f_n\leq 1]}f_n+\varepsilon.$$ Letting $$n\to\infty$$ and applying the dominated convergence theorem then completes the proof.