# Thread: Proving a Sequence is Cauchy proof Help!

1. ## Find Limit of a Function

Let $\displaystyle s_{1}=\sqrt{p},$where$\displaystyle p>0$and

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

2. 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
$\displaystyle t_{n}=1+ \frac{1}{2!} +...+ \frac{1}{n!}$ 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
$\displaystyle |t_{m}-t_{n}|=\frac{1}{n+1!}+\frac{1}{n+2!}+\frac{1}{n+3! }+...+\frac{1}{m-1!}+\frac{1}{m!}$

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 $\displaystyle \epsilon$ for some N. I used a calculator and it appeared that a bound was $\displaystyle \frac{e-1}{n!}$. 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 $\displaystyle 1+\frac{1}{1!}+\frac{1}{2!}+...\frac{1}{n!} = e$ 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 $\displaystyle t_{n+1}-t_n$ and then use the fact that $\displaystyle t_{n+m}-t_n=\sum_{i=n}^{n+m-1}t_{i+1}-t_i$.

As for your edit, the limit they tend to is e. Take a look at it's Taylor expansion.

3. 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
$\displaystyle t_{n}=1+ \frac{1}{2!} +...+ \frac{1}{n!}$ 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
$\displaystyle |t_{m}-t_{n}|=\frac{1}{n+1!}+\frac{1}{n+2!}+\frac{1}{n+3! }+...+\frac{1}{m-1!}+\frac{1}{m!}$

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 $\displaystyle \epsilon$ for some N. I used a calculator and it appeared that a bound was $\displaystyle \frac{e-1}{n!}$. 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 $\displaystyle 1+\frac{1}{1!}+\frac{1}{2!}+...\frac{1}{n!} = e$ if this is in fact true does anyone know where I can find the proof of this?

The ACTUAL fact us that $\displaystyle \sum^\infty_{n=0}\frac{1}{n!}=e$ ...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 $\displaystyle \ln x = 1$ .

Tonio

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

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

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

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

Thus $\displaystyle L = \frac{1+\sqrt{4p+1}}{2}$.

5. ## Cauchy sequence

Originally Posted by tonio
The ACTUAL fact us that $\displaystyle \sum^\infty_{n=0}\frac{1}{n!}=e$ ...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 $\displaystyle \ln x = 1$ .

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