What is the simplest way to proove
$\displaystyle \lim_{n \to \infty} (1 + \frac{1}{n})^n = e$
Thanks
This is the value of $\displaystyle e$ by definition.
Recall that $\displaystyle e$ is defined to be the base of the exponential function such that this function is its own derivative.
So if $\displaystyle \frac{d}{dx}(e^x) = e^x$ then we have
$\displaystyle e^x = \lim_{h \to 0}\frac{e^{x + h} - e^x}{h}$
$\displaystyle e^x = \lim_{h \to 0}\frac{e^xe^h - e^x}{h}$
$\displaystyle e^x = \lim_{h \to 0}\frac{e^x(e^h - 1)}{h}$
$\displaystyle e^x = e^x\lim_{h \to 0}\frac{e^h - 1}{h}$.
Clearly for this equation to hold true, that must mean
$\displaystyle \lim_{h \to 0}\frac{e^h - 1}{h} = 1$
$\displaystyle \lim_{h \to 0}(e^h - 1) = \lim_{h \to 0}h$
$\displaystyle \lim_{h \to 0}e^h - \lim_{h \to 0}1 = \lim_{h \to 0}h$
$\displaystyle \lim_{h \to 0}e^h = \lim_{h \to 0}1 + \lim_{h \to 0}h$
$\displaystyle \lim_{h \to 0}e^h = \lim_{h \to 0}(1 + h)$
$\displaystyle \lim_{h \to 0}(e^h)^{\frac{1}{h}} = \lim_{h \to 0}(1 + h)^{\frac{1}{h}}$
$\displaystyle \lim_{h \to 0}e = \lim_{h \to 0}(1 + h)^{\frac{1}{h}}$
$\displaystyle e = \lim_{h \to 0}(1 + h)^{\frac{1}{h}}$.
Now supposing we let $\displaystyle n = \frac{1}{h}$, clearly as $\displaystyle h \to 0, n \to \infty$, so this means
$\displaystyle e = \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n$.
Q.E.D.
Hello!
First of all e is just a sing for "special" number.
So, what you have is a infinite sequence of numbers.
Theorem:
If infinite sequence of numbers is bounded and monotonic, the our sequence is converges, in other words there is some number which is called the limit of infinite sequence of numbers.
From our theorem we need to prove two things:
1. Boundedness
2. Monotonicity.(?)
Prove: using Inequality of arithmetic and geometric means - Wikipedia, the free encyclopedia
1. $\displaystyle \sqrt[n+1]{(1+\frac{1}{n})^n\cdot 2} < \frac{n+1+2}{n+1}<3$
2. $\displaystyle \sqrt[n+1]{(1+\frac{1}{n})^n\cdot 1}< \frac{1+n+1}{n+1}$
$\displaystyle (1 + \frac{1}{n})^n<(1 + \frac{1}{n+1})^{n+1}$
The correct setting of the problem is the proof that the sequence $\displaystyle a_{n} = (1+\frac{1}{n})^{n}$ converges to a finite number that we call 'e' and is $\displaystyle 2 < e < 3$. All other results, like for example the well known relation $\displaystyle \frac{d}{dx} e^{x} = e^{x}$ used in the 'demostration', are derived from this fundamental preliminary result and not vice-versa...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
Incorrect, $\displaystyle e$ is defined to be the base of the exponential function such that this function is its own derivative.
Therefore the setting of $\displaystyle \frac{d}{dx}(e^x) = e^x$ from the definition and finding an expression for $\displaystyle e$ from the definition is perfectly acceptable.
But yes, you are correct in saying that when you get to the expression you need to prove $\displaystyle \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n$ exists and can be evaluated.
Very well! ... in order to sustain Your thesis You have to demonstrate that is $\displaystyle \frac{d}{dx} e^{x} = e^{x}$ independently by the fact that $\displaystyle \lim_{r \rightarrow \infty} (1+\frac{1}{r})^{r}$ exists finite and it is between 2 and 3...
You are welcome!...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
There are a number of ways of defining "e", since we regard what was to be proven as a definition it is obvious that this is to be proven from some other definition. Two that come to mind immediately are the base of natural logarithms and the solution of the IVP (differential equation).
Earlier posts in this thread have done what is required admirably.
Your post quoted above comes close to being insulting and will only lead to trouble.
CB
Very well...
Assume that you can write $\displaystyle e^x$ as a polynomial $\displaystyle c_0 + c_1x + c_2x^2 + c_3x^3 + \dots$.
Therefore
$\displaystyle e^x = c_0 + c_1x + c_2x^2 + c_3x^3 + \dots$
By setting $\displaystyle x = 0$ we can see $\displaystyle c_0 = 1$.
We also know that $\displaystyle \frac{d}{dx}(e^x) = e^x$ by definition, so differentiating both sides gives...
$\displaystyle e^x = c_1 + 2c_2x + 3c_3x^2 + 4c_4x^3 + \dots$.
Let $\displaystyle x = 0$ and we can see $\displaystyle c_1 = 1$.
Differentiate both sides:
$\displaystyle e^x = 2c_2 + 2\cdot 3c_3x + 3\cdot 4c_4x^2 + 4\cdot 5 c_5x^3 + \dots$.
Let $\displaystyle x = 0$ and we can see $\displaystyle c_2 = \frac{1}{2}$.
Differentiate both sides
$\displaystyle e^x = 2\cdot 3c_3 + 2\cdot 3 \cdot 4 c_4x + 3\cdot 4 \cdot 5c_5 x^2 + 4 \cdot 5 \cdot 6c_6x^3 + \dots$.
Let $\displaystyle x = 0$ and we can see $\displaystyle c_3 = \frac{1}{2\cdot 3}$.
It can be seen that if you were to continue this way you would get
$\displaystyle e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{3\cdot 2} + \frac{x^4}{4\cdot 3\cdot 2} + \frac{x^5}{5\cdot 4 \cdot 3\cdot 2} + \dots$
$\displaystyle e^x = \sum_{r = 0}^{\infty}\frac{x^r}{r!}$.
By letting $\displaystyle x = 1$ we find
$\displaystyle e^1 = \sum_{r = 0}^{\infty}\frac{1}{r!}$.
But we have also shown that $\displaystyle e = \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n$.
Therefore $\displaystyle \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n = \sum_{r = 0}^{\infty}\frac{1}{r!}$, proving that this limit exists, and by evaluating the sum to an acceptable number of terms, can be seen to be $\displaystyle 2.718\,281\,828\,459\,045\,235\,36\dots$.
... only a minor detail [among many other details...] : You were asked to prove that is $\displaystyle \frac{d}{dx} e^{x} = e^{x}$ , so that Your 'answer' that it is true 'by definition' doesn't seem to be fully adequate...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
Maybe I'm dense, but I thought the original question was to prove
$\displaystyle \lim_{n \to \infty} (1 + \frac{1}{n})^n = e$.
The original poster wanted the simplest way to prove this, and he/she did not say anwhere that we could not use the $\displaystyle \frac{d}{dx} e^{x} = e^{x}$ definition. So I don't know why the insistence to prove it without this definition.