Apply Dominated Convergence Theorem: $\lim_{n\rightarrow\infty}\int_0^\infty \dfrac{1+nx^2+n^2x^4}{(1+x^2)^n}d\mu.$ I am working on a problem which I assume is an application of Dominated Convergence Theorem. I am supposed to find
$$\lim_{n\rightarrow\infty}\int_0^\infty \dfrac{1+nx^2+n^2x^4}{(1+x^2)^n}d\mu.$$
I guess I need to find a bound on
$$\frac{1+nx^2+n^2x^4}{(1+x^2)^n}=\frac{1+nx^2}{(1+x^2)^n}+\frac{n^2x^4}{(1+x^2)^n}.$$
I can bound the first part, $ \dfrac{1+nx^2}{(1+x^2)^n}$, by using the binomial expression of ${(1+x^2)^n}$. But I am stuck with dealing with the second part, $\dfrac{n^2x^4}{(1+x^2)^n}$.
Thank you in advance for any suggestions and help.
 A: Presumably $\mu$ is not identically zero Borel measure that is finite on bounded sets of $[0,\infty)$.
As $n^2\sim 2\binom{n}{2}$ as $n\rightarrow\infty$, there is $N$ large enough so that
$$f_n(x):=\frac{1+nx^2+n^2x^4}{(1+x^2)^n}\leq\frac32\frac{1+nx^2+\binom{n}{2}x^4}{(1+x^2)^n}\leq \frac32, \quad x\geq0,n\in\mathbb{N}$$
where the last inequality follows from using binomial expansion in denominator.
Split the integral in two pieces: one over $[0,1]$ and other over $(1,\infty)$.

*

*$\int_{[0,1]}f_n(x)\,\mu(dx)\xrightarrow{n\rightarrow\infty}\int_{[0,1]}\mathbb{1}_{\{0\}}(x)\mu(dx)=\mu(\{0\})$ by dominated convergence.

*Consider the function $\phi(x,u)=\frac{1+ux^2+u^2x^4}{(1+x^2)^u}$ on $(1,\infty)\times(2,\infty)$. Then one can verify that for some $u_0>0$
$$\partial_u\phi(x,u)=(1+x^2)^{-u}\Big(x^2+2ux^4-\log(1+x^2)(1+ux^2+u^2x^4)\big)<0$$
whenever $u\geq u_0$ and $x>1$.
This means that the sequence $\{f_n: n\geq u_0\}$ is decreasing over $(1,\infty)$. Therefore, if $f_m(x)=\frac{1+mx^2+m^2x^4}{(1+x^2)^m}$ is $\mu$-integrable over $(1,\infty)$ form some $m$ large enough, then we can apply dominated convergence to obtain that
$$\lim_n\int_{(1,\infty)}f_n(x)\,\mu(dx)=0$$
Putting things together, we obtain that
$$\lim_n\int_{[0,\infty)}f_n\,d\mu=\mu(\{0\})$$
