# Sequence convergence proof

• July 18th 2010, 01:43 AM
losm1
Sequence convergence proof
What is the simplest way to proove

$\lim_{n \to \infty} (1 + \frac{1}{n})^n = e$

Thanks
• July 18th 2010, 01:57 AM
Prove It
This is the value of $e$ by definition.

Recall that $e$ is defined to be the base of the exponential function such that this function is its own derivative.

So if $\frac{d}{dx}(e^x) = e^x$ then we have

$e^x = \lim_{h \to 0}\frac{e^{x + h} - e^x}{h}$

$e^x = \lim_{h \to 0}\frac{e^xe^h - e^x}{h}$

$e^x = \lim_{h \to 0}\frac{e^x(e^h - 1)}{h}$

$e^x = e^x\lim_{h \to 0}\frac{e^h - 1}{h}$.

Clearly for this equation to hold true, that must mean

$\lim_{h \to 0}\frac{e^h - 1}{h} = 1$

$\lim_{h \to 0}(e^h - 1) = \lim_{h \to 0}h$

$\lim_{h \to 0}e^h - \lim_{h \to 0}1 = \lim_{h \to 0}h$

$\lim_{h \to 0}e^h = \lim_{h \to 0}1 + \lim_{h \to 0}h$

$\lim_{h \to 0}e^h = \lim_{h \to 0}(1 + h)$

$\lim_{h \to 0}(e^h)^{\frac{1}{h}} = \lim_{h \to 0}(1 + h)^{\frac{1}{h}}$

$\lim_{h \to 0}e = \lim_{h \to 0}(1 + h)^{\frac{1}{h}}$

$e = \lim_{h \to 0}(1 + h)^{\frac{1}{h}}$.

Now supposing we let $n = \frac{1}{h}$, clearly as $h \to 0, n \to \infty$, so this means

$e = \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n$.

Q.E.D.
• July 18th 2010, 02:15 AM
Also sprach Zarathustra
Quote:

Originally Posted by losm1
What is the simplest way to proove

$\lim_{n \to \infty} (1 + \frac{1}{n})^n = e$

Thanks

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. $\sqrt[n+1]{(1+\frac{1}{n})^n\cdot 2} < \frac{n+1+2}{n+1}<3$

2. $\sqrt[n+1]{(1+\frac{1}{n})^n\cdot 1}< \frac{1+n+1}{n+1}$

$(1 + \frac{1}{n})^n<(1 + \frac{1}{n+1})^{n+1}$
• July 18th 2010, 04:46 AM
chisigma
The correct setting of the problem is the proof that the sequence $a_{n} = (1+\frac{1}{n})^{n}$ converges to a finite number that we call 'e' and is $2 < e < 3$. All other results, like for example the well known relation $\frac{d}{dx} e^{x} = e^{x}$ used in the 'demostration', are derived from this fundamental preliminary result and not vice-versa...

Kind regards

$\chi$ $\sigma$
• July 18th 2010, 05:01 AM
Prove It
Incorrect, $e$ is defined to be the base of the exponential function such that this function is its own derivative.

Therefore the setting of $\frac{d}{dx}(e^x) = e^x$ from the definition and finding an expression for $e$ from the definition is perfectly acceptable.

But yes, you are correct in saying that when you get to the expression you need to prove $\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n$ exists and can be evaluated.
• July 18th 2010, 05:35 AM
chisigma
Quote:

Originally Posted by Prove It
Incorrect, $e$ is defined to be the base of the exponential function such that this function is its own derivative... therefore the setting of $\frac{d}{dx}(e^x) = e^x$ from the definition and finding an expression for $e$ from the definition is perfectly acceptable...

Very well! ... in order to sustain Your thesis You have to demonstrate that is $\frac{d}{dx} e^{x} = e^{x}$ independently by the fact that $\lim_{r \rightarrow \infty} (1+\frac{1}{r})^{r}$ exists finite and it is between 2 and 3...

You are welcome!...

Kind regards

$\chi$ $\sigma$
• July 18th 2010, 06:13 AM
CaptainBlack
Quote:

Originally Posted by Prove It
Incorrect, $e$ is defined to be the base of the exponential function such that this function is its own derivative.

Therefore the setting of $\frac{d}{dx}(e^x) = e^x$ from the definition and finding an expression for $e$ from the definition is perfectly acceptable.

But yes, you are correct in saying that when you get to the expression you need to prove $\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n$ exists and can be evaluated.

When replying to a post please quote the post you are responding to, it avoids confusion.

CB
• July 18th 2010, 06:17 AM
CaptainBlack
Quote:

Originally Posted by chisigma
Very well! ... in order to sustain Your thesis You have to demonstrate that is $\frac{d}{dx} e^{x} = e^{x}$ independently by the fact that $\lim_{r \rightarrow \infty} (1+\frac{1}{r})^{r}$ exists finite and it is between 2 and 3...

You are welcome!...

Kind regards

$\chi$ $\sigma$

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).

Your post quoted above comes close to being insulting and will only lead to trouble.

CB
• July 18th 2010, 06:21 AM
Prove It
Very well...

Assume that you can write $e^x$ as a polynomial $c_0 + c_1x + c_2x^2 + c_3x^3 + \dots$.

Therefore

$e^x = c_0 + c_1x + c_2x^2 + c_3x^3 + \dots$

By setting $x = 0$ we can see $c_0 = 1$.

We also know that $\frac{d}{dx}(e^x) = e^x$ by definition, so differentiating both sides gives...

$e^x = c_1 + 2c_2x + 3c_3x^2 + 4c_4x^3 + \dots$.

Let $x = 0$ and we can see $c_1 = 1$.

Differentiate both sides:

$e^x = 2c_2 + 2\cdot 3c_3x + 3\cdot 4c_4x^2 + 4\cdot 5 c_5x^3 + \dots$.

Let $x = 0$ and we can see $c_2 = \frac{1}{2}$.

Differentiate both sides

$e^x = 2\cdot 3c_3 + 2\cdot 3 \cdot 4 c_4x + 3\cdot 4 \cdot 5c_5 x^2 + 4 \cdot 5 \cdot 6c_6x^3 + \dots$.

Let $x = 0$ and we can see $c_3 = \frac{1}{2\cdot 3}$.

It can be seen that if you were to continue this way you would get

$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{3\cdot 2} + \frac{x^4}{4\cdot 3\cdot 2} + \frac{x^5}{5\cdot 4 \cdot 3\cdot 2} + \dots$

$e^x = \sum_{r = 0}^{\infty}\frac{x^r}{r!}$.

By letting $x = 1$ we find

$e^1 = \sum_{r = 0}^{\infty}\frac{1}{r!}$.

But we have also shown that $e = \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n$.

Therefore $\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n = \sum_{r = 0}^{\infty}\frac{1}{r!}$, proving that this limit exists, and by evaluating the sum to an acceptable number of terms, can be seen to be $2.718\,281\,828\,459\,045\,235\,36\dots$.
• July 18th 2010, 06:43 PM
chisigma
Quote:

Originally Posted by Prove It

... we also know that $\frac{d}{dx}(e^x) = e^x$ by definition...

... only a minor detail [among many other details...] : You were asked to prove that is $\frac{d}{dx} e^{x} = e^{x}$ , so that Your 'answer' that it is true 'by definition' doesn't seem to be fully adequate...

Kind regards

$\chi$ $\sigma$
• July 18th 2010, 07:21 PM
eumyang
Quote:

Originally Posted by chisigma
... only a minor detail [among many other details...] : You were asked to prove that is $\frac{d}{dx} e^{x} = e^{x}$ , so that Your 'answer' that it is true 'by definition' doesn't seem to be fully adequate...

Maybe I'm dense, but I thought the original question was to prove
$\lim_{n \to \infty} (1 + \frac{1}{n})^n = e$.

The original poster wanted the simplest way to prove this, and he/she did not say anwhere that we could not use the $\frac{d}{dx} e^{x} = e^{x}$ definition. So I don't know why the insistence to prove it without this definition.
• July 18th 2010, 08:12 PM
Prove It
Quote:

Originally Posted by chisigma
... only a minor detail [among many other details...] : You were asked to prove that is $\frac{d}{dx} e^{x} = e^{x}$ , so that Your 'answer' that it is true 'by definition' doesn't seem to be fully adequate...

Kind regards

$\chi$ $\sigma$

Incorrect once again. The question was asking to prove $\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n = e$.

Using the definition $\frac{d}{dx}(e^x) = e^x$ is once again appropriate.