# prove that a sequence converges

• Apr 30th 2010, 02:18 PM
santiagos11
prove that a sequence converges
Prove that the sequence $\displaystyle x_{n}=n^{1/n}$ converges.
I know it converges to 1. We need to use the definition of sequence convergence: $\displaystyle x_{n} \rightarrow L \Leftrightarrow(\forall\epsilon>0)(\exists N\in \mathbb{N})(\forall n\in \mathbb{N})(n>N \Rightarrow |x_{n}-L|<\epsilon)$
Another thing, obviously $\displaystyle |x_{n}-L|=|n^{1/n}-1|=n^{1/n}-1$ since $\displaystyle n^{1/n}-1\geq0$
• Apr 30th 2010, 02:58 PM
Plato
Quote:

Originally Posted by santiagos11
Prove that the sequence x(n)=n^(1/n) converges.

Let $\displaystyle y_n=\sqrt[n]{n} - 1$ so $\displaystyle y_n\ge 0$ and
by the binominal theorem $\displaystyle n=(y_n+1)^n>\frac{n(n-1)}{2}y_n^2$.
Therefore $\displaystyle 0\le y_n<\sqrt{\frac{2}{n-1}}$.
• Apr 30th 2010, 03:36 PM
santiagos11
Well, you are certanly right. But how does that fit the standard definition?
• Apr 30th 2010, 04:02 PM
Plato
Quote:

Originally Posted by santiagos11
Well, you are certanly right. But how does that fit the standard definition?

Having taught this 'stuf' for thirty+ years I have never seen a $\displaystyle \epsilon\-\delta$ proof given for this limit.
So I can't imagine what you mean be 'standard'.
• Apr 30th 2010, 05:04 PM
santiagos11
For example, here is a partial proof:
Let $\displaystyle \epsilon>0$. Clearly $\displaystyle 1\leq n^{1/n}\leq 2 \ \forall n\in \mathbb{N}$
case 1: $\displaystyle \epsilon>1$
Let $\displaystyle N=1$. Suppose $\displaystyle n>N$. Then $\displaystyle |x_{n}-1|=|n^{1/n}-1|=n^{1/n}-1$. On the other hand $\displaystyle \epsilon + 1 > 2 \geq n^{1/n}$, so $\displaystyle n^{1/n}-1<\epsilon$. Therefore, $\displaystyle |x_{n}-1|<\epsilon$.
Do you see how this follows the definition that I gave in the top?
Now it remains to show the same, in the case $\displaystyle 0<\epsilon\leq1$
• Apr 30th 2010, 06:15 PM
Plato
Quote:

Originally Posted by santiagos11
For example, here is a partial proof:
Let $\displaystyle \epsilon>0$. Clearly $\displaystyle 1\leq n^{1/n}\leq 2 \ \forall n\in \mathbb{N}$
case 1: $\displaystyle \epsilon>1$
Let $\displaystyle N=1$. Suppose $\displaystyle n>N$. Then $\displaystyle |x_{n}-1|=|n^{1/n}-1|=n^{1/n}-1$. On the other hand $\displaystyle \epsilon + 1 > 2 \geq n^{1/n}$, so $\displaystyle n^{1/n}-1<\epsilon$. Therefore, $\displaystyle |x_{n}-1|<\epsilon$.
Do you see how this follows the definition that I gave in the top?
Now it remains to show the same, in the case $\displaystyle 0<\epsilon\leq1$

That is a waste of effort. There is absolutely no reason to reinvent the wheel.
The object of proof at this level is to deepen understanding, not to show off.
• May 1st 2010, 06:48 PM
santiagos11
Well yeach I gues you are right. However I do have another problem:
Prove that that the following sequence,defined recursively, converges or diverges:
$\displaystyle x_{1}=1$
$\displaystyle x_{n}=\sqrt{1+x_{n-1}}$