# Proving a Sequence is Cauchy proof Help!

• Mar 21st 2010, 07:19 PM
fizzle45
Find Limit of a Function
Let $s_{1}=\sqrt{p},$where $p>0$and

$s_{n+1}=\sqrt{p+s_n}
$
for n in naturals
Find the limit of $s_{n}$ Hint: One upper bound of the sequence is $1+2\sqrt{p}$
• Mar 21st 2010, 07:38 PM
Focus
Quote:

Originally Posted by fizzle45
Hi, I've been scratching my brain trying to figure out this problem.

Show by using the definition that the following sequence is cauchy
$t_{n}=1+ \frac{1}{2!} +...+ \frac{1}{n!}$ for n belonging to naturals.

I'm fairly new at Latex so I'm going to skip trying to re type all my steps with it but basically i have
$|t_{m}-t_{n}|=\frac{1}{n+1!}+\frac{1}{n+2!}+\frac{1}{n+3! }+...+\frac{1}{m-1!}+\frac{1}{m!}$

Here's where I get a little stuck. I know that my goal is to find a bound for this which will be less than $\epsilon$ for some N. I used a calculator and it appeared that a bound was $\frac{e-1}{n!}$. But I know i just can't claim something is a bound by basis of calculator values, so I need some desperate help on trying to find a bound for this thing. thanks in advance

EDIT:
I just read that $1+\frac{1}{1!}+\frac{1}{2!}+...\frac{1}{n!} = e$ if this is in fact true does anyone know where I can find the proof of this?

The trick on these sorts of questions is to consider $t_{n+1}-t_n$ and then use the fact that $t_{n+m}-t_n=\sum_{i=n}^{n+m-1}t_{i+1}-t_i$.

As for your edit, the limit they tend to is e. Take a look at it's Taylor expansion.
• Mar 21st 2010, 07:41 PM
tonio
Quote:

Originally Posted by fizzle45
Hi, I've been scratching my brain trying to figure out this problem.

Show by using the definition that the following sequence is cauchy
$t_{n}=1+ \frac{1}{2!} +...+ \frac{1}{n!}$ for n belonging to naturals.

I'm fairly new at Latex so I'm going to skip trying to re type all my steps with it but basically i have
$|t_{m}-t_{n}|=\frac{1}{n+1!}+\frac{1}{n+2!}+\frac{1}{n+3! }+...+\frac{1}{m-1!}+\frac{1}{m!}$

Here's where I get a little stuck. I know that my goal is to find a bound for this which will be less than $\epsilon$ for some N. I used a calculator and it appeared that a bound was $\frac{e-1}{n!}$. But I know i just can't claim something is a bound by basis of calculator values, so I need some desperate help on trying to find a bound for this thing. thanks in advance

EDIT:
I just read that $1+\frac{1}{1!}+\frac{1}{2!}+...\frac{1}{n!} = e$ if this is in fact true does anyone know where I can find the proof of this?

The ACTUAL fact us that $\sum^\infty_{n=0}\frac{1}{n!}=e$ ...the sum is infinite, and either one defines the number e that way, or else defines some other way and proves the above sum is e. Anyway, any decent calculus book must include a discussion on this. For example, Apostol's "Calculus I" has several interesting parts on the number e, including its definition as that number x for which $\ln x = 1$ .

Tonio
• Mar 21st 2010, 08:14 PM
chiph588@
Quote:

Originally Posted by fizzle45
Let $s_{1}=\sqrt{p},$where $p>0$and

$s_{n+1}=\sqrt{p+s_n}
$
for n in naturals
Find the limit of $s_{n}$ Hint: One upper bound of the sequence is $1+2\sqrt{p}$

Let $L$ be the limit of the sequence $s_n$.

Since $s_{n}^2=p+s_{n-1}$ and for large enough $N\; |s_n-s_{n-1}|<\epsilon$, $L^2=p+L$.

Hence $L$ is the positive solution to $x^2-x-p=0$.

Thus $L = \frac{1+\sqrt{4p+1}}{2}$.
• Mar 22nd 2010, 03:58 AM
kompik
Cauchy sequence
Quote:

Originally Posted by tonio
The ACTUAL fact us that $\sum^\infty_{n=0}\frac{1}{n!}=e$ ...the sum is infinite, and either one defines the number e that way, or else defines some other way and proves the above sum is e. Anyway, any decent calculus book must include a discussion on this. For example, Apostol's "Calculus I" has several interesting parts on the number e, including its definition as that number x for which $\ln x = 1$ .

Tonio

As far as the original problem of showing that $\sum_{k=0}^n \frac{1}{n!}$ is Cauchy, you could try to use
$n!\ge 2^{n-1}$
$\frac1{n!}\le \frac1{2^{n-1}}$