How to show that the sequence $\int_{0}^{\infty} \frac{e^{-nx}}{\sqrt{x}} \,dx$ converges to $0$?

Question

I need to show that for any $\epsilon>0$ exist a $N \in \mathbb{N}$ s.t $n \geq N\in \mathbb{N}$$ \implies$ $\left|\int_{0}^{\infty} \frac{e^{-nx}}{\sqrt{x}} \,dx \right|<\epsilon$.

I know that: \begin{align*} \left|\int_{0}^{\infty} \frac{e^{-nx}}{\sqrt{x}} \,dx \right|\leq \int_{0}^{\infty} \left| \frac{e^{-nx}}{\sqrt{x}} \right| \,dx=\int_{0}^{\infty} \frac{e^{-nx}}{\sqrt{x}} \,dx \end{align*}

But i can't find a function $g(x,n)$ such that:

\begin{align*} \int_{0}^{\infty} \frac{e^{-nx}}{\sqrt{x}} \,dx \leq \int_{0}^{\infty} g(x,n) \,dx \end{align*}

I apreciate your help.

Answer 1

Note that for $n\ge 1$, $\frac{e^{-nx}}{\sqrt x}\le \frac{e^{-x}}{\sqrt x}$ and $\int_0^\infty \frac{e^{-x}}{\sqrt x}\,dx<\infty$. So, we can apply the Dominated Convergence Theorem and find the limit of interest is $0$

However, we don't need the Dominated Convergence Theorem to proceed.

Simply enforcing the substitution $x\mapsto x/n$ reveals

$$\int_0^\infty \frac{e^{-nx}}{\sqrt x}\,dx=\frac1{\sqrt n}\int_0^\infty \frac{e^{-x}}{\sqrt x}\,dx$$

Inasmuch as $\int_0^\infty \frac{e^{-x}}{\sqrt x}\,dx$ exists, then we see that

$$\lim_{n\to \infty}\frac1{\sqrt n}\int_0^\infty \frac{e^{-x}}{\sqrt x}\,dx=0$$

Answer 2

Notice that when $\delta$ is sufficiently close to $0$, the integration $\displaystyle\int_0^\delta\frac1{\sqrt{x}}\;\mathrm{d}x$ converges to $0$.

Also, when $n\to+\infty$, for a fixed $\delta > 0$, $\displaystyle\int_\delta^{+\infty}\frac{e^{-nx}}{\sqrt{x}}\;\mathrm{d}x \leq \frac1{\sqrt\delta}\int_\delta^{+\infty}e^{-nx}\;\mathrm{d}x = \frac1{n\sqrt\delta}e^{-n\delta}$ converges to $0$ as well.

Now given the $\epsilon > 0$, choose the split point $\displaystyle\delta = \frac{\epsilon^2}{16}$. The estimation runs as follows: \begin{align} \int_0^{+\infty}\frac{e^{-nx}}{\sqrt{x}}\;\mathrm{d}x &= \int_0^{\delta}\frac{e^{-nx}}{\sqrt{x}}\;\mathrm{d}x + \int_\delta^{+\infty}\frac{e^{-nx}}{\sqrt{x}}\;\mathrm{d}x \\ &\leq \int_0^{\delta}\frac1{\sqrt{x}}\;\mathrm{d}x + \int_\delta^{+\infty}\frac{e^{-nx}}{\sqrt{x}}\;\mathrm{d}x \\ &=\epsilon/2 + \int_\delta^{+\infty}\frac{e^{-nx}}{\sqrt{x}}\;\mathrm{d}x. \end{align} The only thing left is to pick a $n$ which is sufficiently large so that $\displaystyle\int_\delta^{+\infty}\frac{e^{-nx}}{\sqrt{x}}\;\mathrm{d}x < \epsilon/2$