# Application on the limit definition of e

1. Oct 27, 2013

### fblues

Application on the limit definition of "e"

Hi, I have known that:
(i)$(1+\frac{a}{n})^n=((1+\frac{a}{n})^\frac{n}{a})^a\to e^a$
(ii)$(1-\frac{1}{n})^n=(\frac{n-1}{n})^n=(\frac{1}{\frac{n}{n-1}})^{(n-1)+1}=(\frac{1}{1+\frac{1}{n-1}})^{(n-1)}\cdot (\frac{1}{1+\frac{1}{n-1}}) \to \frac{1}{e}\cdot 1$

With above two facts, I wanted to show $(\frac{1}{1-\frac{t}{\sqrt{\frac{n}{2}}}})^\frac{n}{2} \to e^{\sqrt{\frac{n}{2}}t}\cdot e^\frac{t^2}{2}$ as n goes to infinity, for a fixed positive real t.

However, I am continuously getting $e^{\sqrt{\frac{n}{2}}t}\cdot e^{t^2}$ instead of above result and could not find the reason on the following my argument:

$(\frac{\sqrt{\frac{n}{2}}}{\sqrt{\frac{n}{2}}-t})^\frac{n}{2}=(\frac{(\frac{\sqrt{\frac{n}{2}}}{t}-1)+1}{\frac{\sqrt{\frac{n}{2}}}{t}-1})^\frac{n}{2}=(1+\frac{1}{\frac{\sqrt{\frac{n}{2}}}{t}-1})^{(\frac{\sqrt{\frac{n}{2}}}{t}-1)\sqrt{\frac{n}{2}}t+\sqrt{\frac{n}{2}}t}=(1+\frac{1}{\frac{\sqrt{\frac{n}{2}}}{t}-1})^{(\frac{\sqrt{\frac{n}{2}}}{t}-1)\sqrt{\frac{n}{2}}t}\cdot (1+\frac{1}{\frac{\sqrt{\frac{n}{2}}}{t}-1})^{(\frac{\sqrt{\frac{n}{2}}}{t}-1)t^2+t^2}$
$=(1+\frac{1}{\frac{\sqrt{\frac{n}{2}}}{t}-1})^{(\frac{\sqrt{\frac{n}{2}}}{t}-1)\sqrt{\frac{n}{2}}t}\cdot (1+\frac{1}{\frac{\sqrt{\frac{n}{2}}}{t}-1})^{(\frac{\sqrt{\frac{n}{2}}}{t}-1)t^2}\cdot (1+\frac{1}{\frac{\sqrt{\frac{n}{2}}}{t}-1})^{t^2} \to e^{\sqrt{\frac{n}{2}}t}\cdot e^{t^2}\cdot 1$ as n goes to infinity.

It would be very appreciative if you let me know my mistake.
Thank you very much.

2. Oct 28, 2013

### CompuChip

Hi fblues. How can there be an $n$ on the right hand side, after you have taken the limit of $n \to \infty$?

3. Oct 28, 2013

### Office_Shredder

Staff Emeritus
I think the statement that you want is something like

$$\lim_{n\to \infty} \left(\frac{1}{1-t/\sqrt{n/2}} \right)^{n/2} e^{-\sqrt{n/2} t} = e^{t^2/2}$$

(I don't know if this is the correct statement, but is what your statement should look similar to).

4. Oct 28, 2013

### fblues

To. CompuChip:
Thank you for letting me know. I tried to split the part that I don't know from the original problem and made a mistake during this procedure. BTW, it seems Office_Shredder knows the original one.

To. Office_Shredder:
Yes. The problem is from "the mgf of Chi_sq(n) becomes the mgf of normal(0,1) as n goes to infinity." I think the general approach is use of Taylor expansion. But, I tried to employ the limit definition of e. Do you have an idea for this?