Dominated Convergence Is there a function that dominates
$$f_n(x) = \frac{1}{(1+\frac{x}{n})^nx^{\frac{1}{n}}}$$ 
on $(1,\infty)$ for all $n$? I need to apply DCT to get $e^{-x}$.Obviously MCT doesn't apply since $x^{\frac{1}{n}}$ is decreasing but $(1+\frac{x}{n})^n$ is increasing to $e^x$. The best I got was $\frac{1}{1+x}$ but it's not in $L^1$.
 A: When $n \geq 2$, notice that for $x > 0$ we have
$$ \left( 1 + \frac{x}{n} \right)^n = 1 + x + \frac{n-1}{2n}x^2 + \cdots \geq 1 + x + \frac{1}{4}x^2. $$
Using this we can bound $(f_n)$ on $(1, \infty)$ for $n \geq 2$ by
$$ f_n(x) \leq \frac{1}{1+x+x^2/4}. $$
A: First, we have for $x\in[1,\infty)$
$$x^{-1/n}\le 1 \tag 1$$
Next, we show that $\left(1+\frac xn\right)^n$ is increasing function of $n$ for $x>0$.  To that end, we analyze the ratio
$$\begin{align}
\frac{\left(1+\frac x{n+1}\right)^{n+1}}{\left(1+\frac xn\right)^n}&=\left(\frac{n(n+1+x)}{(n+1)(n+x)}\right)^{n+1}\left(1+\frac xn\right)\\\\
&=\left(1-\frac{x}{(n+1)(n+x)}\right)^{n+1}\left(1+\frac xn\right) \tag 2\\\\
&\ge \left(1-\frac{x}{n+x}\right)\left(1+\frac xn\right) \tag 3\\\\
&=1
\end{align}$$
where we used Bernoulli's Inequality to go from $(2)$ to $(3)$.  We have shown that $\left(1+\frac xn\right)^n$ is increasing in $n$.  Therefore, we can assert for $n\ge 2$
$$\begin{align}
\frac{1}{\left(1+\frac xn\right)^n}&\le \frac{1}{\left(1+\frac x2\right)^2}\\\\
&=\frac{1}{1+x+\frac14x^2}\tag 4
\end{align}$$
Putting all of this together, we have from $(1)$ and $(4)$
$$\frac{1}{\left(1+\frac xn\right)^nx^{1/n}}\le \frac{1}{1+x+\frac14 x^2}$$
Inasmuch as the integral of the dominating function satisfies the condition
$$\int_1^{\infty}\frac{1}{1+x+\frac14 x^2}\,dx<\infty$$
then by the Dominated Convergence Theorem, we have
$$\begin{align}
\lim_{n\to \infty}\int_1^{\infty}\frac{1}{\left(1+\frac xn\right)^nx^{1/n}}\,dx&=\int_1^{\infty}\lim_{n\to \infty}\left(\frac{1}{\left(1+\frac xn\right)^nx^{1/n}}\right)\,dx\\\\
&=\int_1^\infty e^{-x}\,dx\\\\
&=1
\end{align}$$
