1. ## def of sequence

Use the definition of convergent to prove :
$\displaystyle {(2n)}^{\frac {1}{n}} \to 1$

2. Originally Posted by flower3
Use the definition of convergent to prove :
$\displaystyle {(2n)}^{\frac {1}{n}} \to 1$
Yes, what have you done? What is the definition of convergence for a sequence?

For this particular problem, you might consider taking the logarithm of every number in the sequence and looking at that new sequence.

3. $\displaystyle proof:$
$\displaystyle let \ \epsilon >0 \ want \ to \ prove \ \exists k \in N \ such \ that \ \mid (2n)^{ \frac {1}{n}} \ - 1 \mid < \epsilon ,\ \forall \ n \geq k$
then??

4. $\displaystyle (2n)^{1/n} = 2^{1/n} n^{1/n}$. Obviously $\displaystyle 2^{1/n} \rightarrow 1$, so we need to only prove that $\displaystyle n^{1/n} \rightarrow 1$.

For n > 1 we have $\displaystyle n^{1/n} > 1^{1/n} = 1$. So we can let $\displaystyle n^{1/n} = 1 + h_n$ where $\displaystyle h_n$ is some positive quantity depending on n.

So $\displaystyle n = (1 + h_n)^n$. Now, it is clear from the binomial theorem that $\displaystyle (1 + h_n)^n \ge 1 + nh_n + \frac{n(n+1)}{2} h_n \ge \frac{n(n+1)}{2} h_n$.

Can you show that $\displaystyle h_n$ goes to 0 and so $\displaystyle n^{1/n} = 1 + h_n$ goes to 1?