Aproximation of the Normal Distribution by the Normal Density Function

Question

In Feller's introduction to probability the next lemma is stated:

"As $x\rightarrow \infty$ $\tag1 \frac{1-R(x)}{x^{-1}n(x)} \rightarrow 1$

Where $R(x)$ is the normal distribution and $n(x)$ is the normal density function. And more precisely:

$\tag1 (x^{-1}-x^{-3})n(x) < 1-R(x) < x^{-1}n(x)$ "

This lemma is shown by proving that the left and right hand side of the inequality are the integrals from $x$ to $\infty$ of $(1-3x^{-4})n(x)$ and $((1+x^{-2})n(x))$ and using the inequality: $\tag1(1-3x^{-4})n(x) < n(x) < (1+x^{-2})n(x)$

The problem i have is in a generalization of this that comes as a problem of the book in the same chapter. The statement of the problem is:

"Prove that: $\tag1 \frac{1-R(x)}{\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}(\frac{1}{x}-\frac{1}{x^3}+\cdots+(-1)^k\frac{1\cdot3\cdot7\cdots(2k-1)}{x^{(2k+1)}})}\rightarrow 1$ for $x>0$".

I've only tried to show that a similar inequality like the second one holds like in the lemma using the derivatives of $x^{-k}n(x)$ and $(x^{-k}-x^{-(k+2)})n(x)$ but i don't see exactly how to get the sum that is involved, so any help would be much appreciated.

Answer 1

About the statement

$$ \frac{1-R(x)}{\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}(\frac{1}{x}-\frac{1}{x^3}+...+(-1)^k\frac{1*3*7*...*(2k-1)}{x^{(2k+1)}})}\xrightarrow{x \to +\infty} 1$$

In fact, it holds true, even for the general form like this one

$$ \frac{1-R(x)}{\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}(\frac{1}{x}+\sum_{k=2}^n \frac{a_k}{x^k})}\xrightarrow{x \to +\infty} 1 \qquad \forall n\ge 2, \forall (a_k)_{k=1,...,n} \in \mathbb{R}$$

The reason is $$\frac{\frac{1}{x}+\sum_{k=2}^n \frac{a_k}{x^k}}{\frac{1}{x}}=1+\sum_{k=2}^n \frac{a_k}{x^{k-1}}\xrightarrow{x \to +\infty} 1$$ Then, you apply the lemma in Fella's book:

$$\frac{1-R(x)}{\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}(\frac{1}{x}+\sum_{k=2}^n \frac{a_k}{x^k})} = \frac{1-R(x)}{\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}(\frac{1}{x})}\times \frac{\frac{1}{x}+\sum_{k=2}^n \frac{a_k}{x^k}}{\frac{1}{x}} \xrightarrow{x \to +\infty} 1$$

PS: I provide another proof for Feller's lemma using the Hopital's rule

Proof: $$\frac{1-R(x)}{x^{-1}n(x)} = \frac{\int_x^{+\infty}\exp\left(-\frac{t^2}{2} \right)dt}{\frac{1}{x}e^{-\frac{1}{2}x^2}}$$ Let's denote $$G(x) = \int_x^{+\infty}e^{-\frac{1}{2}t^2}dt$$ $$F(x) =\frac{1}{x}e^{-\frac{1}{2}x^2} $$ We have $$\lim_{x \to +\infty}G(x) = 0$$ $$\lim_{x \to +\infty}F(x) = 0$$ Apply the Hopital's rule for $c = +\infty$ \begin{align} \lim_{x \to +\infty}\frac{G(x)}{F(x)} &= \lim_{x \to +\infty}\frac{G(x)}{F(x)}\\ &= \lim_{x \to +\infty}\frac{G'(x)}{F'(x)}\\ &= \lim_{x \to +\infty}\frac{-e^{-\frac{1}{2}x^2}}{-e^{-\frac{1}{2}x^2 \left(1+ \frac{1}{x^2} \right)}} &= \lim_{x \to +\infty}\frac{1}{1+ \frac{1}{x^2} } \\ &= 1 \end{align} Q.E.D