# Thread: prove that a sequence converges

1. ## prove that a sequence converges

Prove that the sequence $x_{n}=n^{1/n}$ converges.
I know it converges to 1. We need to use the definition of sequence convergence: $x_{n} \rightarrow L \Leftrightarrow(\forall\epsilon>0)(\exists N\in \mathbb{N})(\forall n\in \mathbb{N})(n>N \Rightarrow |x_{n}-L|<\epsilon)$
Another thing, obviously $|x_{n}-L|=|n^{1/n}-1|=n^{1/n}-1$ since $n^{1/n}-1\geq0$

2. Originally Posted by santiagos11
Prove that the sequence x(n)=n^(1/n) converges.
Let $y_n=\sqrt[n]{n} - 1$ so $y_n\ge 0$ and
by the binominal theorem $n=(y_n+1)^n>\frac{n(n-1)}{2}y_n^2$.
Therefore $0\le y_n<\sqrt{\frac{2}{n-1}}$.

3. Well, you are certanly right. But how does that fit the standard definition?

4. 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 $\epsilon\-\delta$ proof given for this limit.
So I can't imagine what you mean be 'standard'.

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

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

7. Well yeach I gues you are right. However I do have another problem:
Prove that that the following sequence,defined recursively, converges or diverges:
$x_{1}=1$
$x_{n}=\sqrt{1+x_{n-1}}$