Dominated convergence theorem with $f_n(x)=\frac{1}{\sqrt{2\pi}}e^{-\sqrt{n}x}\left(1+\frac{x}{\sqrt{n}}\right)^n\chi_{[-\sqrt{n},+\infty]}$ In my textbook there is an exercise where I have to apply the dominated convergence theorem on the function :
$$f_n(x)=\frac{1}{\sqrt{2\pi}}e^{-\sqrt{n}x}\left(1+\frac{x}{\sqrt{n}}\right)^n\chi_{[-\sqrt{n},+\infty[}$$
I can prove that $f_n$ pointwise converges to $\frac{e^{-x^2/2}}{\sqrt{2\pi}}$. Indeed $\left(1+\frac{x}{\sqrt{n}}\right)^n=\exp(n\ln(1+x/\sqrt{n}))= \exp(n(\frac{x}{\sqrt n}-\frac{x^2}{2n})+o(1))$, so $e^{-\sqrt{n}x}\left(1+\frac{x}{\sqrt{n}}\right)^n\rightarrow e^{-x^2/2}$, hence the result.
But I cant manage to dominate the $f_n$ with an integrable function, apparently I should be able to dominate them with $e^{-x^2/3}$, but I can't find how.
Progress : I know that $e^{-\sqrt{n}x}\chi_{[-\sqrt{n},+\infty]}\le e^{x^2}$, because $-\sqrt{n}\le x$ on $[-\sqrt{n},+\infty[$. But I don't see how to dominate the $\left(1+\frac{x}{\sqrt{n}}\right)^n$ part.
 A: You can't expect $e^{-x^2/3}$ to be a dominating function, as that decays way faster than any $f_n$ at $\infty.$  I'm not sure how bounding by $e^{x^2}$ can help, as that is far from being in $L^1.$
Hint: Look for a dominating function of the form $Ce^{a|x|},$ where $a$ is a negative constant. (I think I've been able to do it with $a = (4/3)\ln (1 +(3/4))-1,$ but I need to check my work.)

Added later: I'll be ignoring the $1/\sqrt {2\pi}$ throughout. Start with
$$\tag 1 \ln f_n(x) = -\sqrt n x+ n \ln(1+x/\sqrt n).$$
For $-1 <u \le 0,$ $\ln (1+u) \le u - u^2/2.$ Thus for $x\in (-\sqrt n,0],$ $(1) \le -x^2/2.$ It follows that for this range of $x,$ $f_n(x) \le e^{-x^2/2}.$
Now for $u\ge 0,$ $\ln (1+u) \le u - u^2/2 + u^3/3.$ This leads us to the claim: If $x\in [0,3\sqrt n/4],$ then
$$(1) \le -x^2/2 + x^3/(3\sqrt n) \le -x^2/4.$$
That's easy to verify. So for $x\in [0,3\sqrt n/4],$ $f_n(x) \le e^{-x^2/4}.$
Finally, on $[3\sqrt n/4,\infty),$ we can use the fact that $[\ln (1+u)]/u$ strictly decreases from $1$ to $0$ on $[0,\infty).$ Rewriting $(1)$ gives the expression
$$\tag 2  -\sqrt n x + \sqrt n x\frac{\ln(1+x/\sqrt n)}{x/\sqrt n}.$$
For $x\ge 3\sqrt n/4,$ the fraction in $(2)$ is bounded above by $[\ln(1+(3/4)]/(3/4) < 1.$ Letting $a = [\ln(1+(3/4))]/(3/4)-1,$ we see $(2)$ is bounded above by $a\sqrt n x \le ax.$ Thus on $[3\sqrt n/4,\infty),$ $f_n \le e^{ax}.$
Let's now add up these dominating functions. We get, for each $n,$
$$f_n(x) \le e^{-x^2/2} + e^{-x^2/4} + e^{a|x|}$$
for $x\in \mathbb R.$ That's enough to invoke the dominated convergence theorem, but if you want a simpler answer, note that the first two functions are bounded by constant multiples of $e^{a|x|},$ so some $Ce^{a|x|}$ will work.
