# Binomial expansion

1. Jan 10, 2005

### recon

How do you expand and simplify $$(1 + \sqrt\frac{2}{n-1})^n$$?

I know this involves a binomial expansion and I can expand it to look something like

$$\left(\begin{array}{c}n&0\end{array}\right){\frac{2}{n-1}}^\frac{0}{2} + \left(\begin{array}{c}n&1\end{array}\right){\frac{2}{n-1}}^\frac{1}{2} + ...$$

but how do you simplify this?

Last edited: Jan 10, 2005
2. Jan 10, 2005

### dextercioby

Sorry to dissapoint you,but apparently u cannot.It looks kinda ugly,but that's how it was supposed to be,since it involved a radical.

Daniel.

3. Jan 11, 2005

### recon

Hmm. How can I solve the following problem then?

n is a positive integer. Prove that:

$$n^\frac{1}{n} < 1 + \sqrt\frac{2}{n-1}$$

I also need to show that $$n^\frac{1}{n} \rightarrow 1$$ as $$n \rightarrow \infty$$. I know this follows logically from the fact that $$\frac{1}{n} \rightarrow 0$$as $$n \rightarrow \infty$$. Is there a more rigorous way for showing this?

Also, what is the maximum value of $$n^\frac{1}{n}$$?

4. Jan 11, 2005

### dextercioby

Do you know calculus??If u did,then
$$\lim_{n\rightarrow +\infty} n^{\frac{1}{n}}=\alpha$$(1)
U need to show that \alpha=1.
Take natural logarithm from both sides.Then
$$\lim_{n\rightarrow +\infty} \frac{1}{n}\ln n =\ln\alpha$$ (2)

The first limit is zero (you can show that considering the function [\itex] \frac{\ln x}{x} [/itex] and using L'Ho^spital rule.
THerefore $\ln\alpha=0 \Rightarrow \alpha=1$.

Daniel.

5. Jan 11, 2005

### recon

I've never studied L'Hospital's Rule before (I just finished Grade 10). However, I just looked it up on the internet, and I do understand how it works, but not why it works.

Is this problem solvable?

If n is a positive integer, prove that:
$$n^\frac{1}{n} < 1 + \sqrt\frac{2}{n-1}$$

6. Jan 11, 2005

### NateTG

L'Hospitals rule is relatively easy to prove using the definitions of limit and derivative.

If n=1 then there are some problems with this.
For n bigger than 1, you've almost got the proof.

Here's something you might find useful:
$$\left(\begin{array}{c}a&b\end{array}\right) = \frac{a!}{b!(a-b)!}$$
Specifically
$$\left(\begin{array}{c}n&0\end{array}\right) = 1$$
$$\left(\begin{array}{c}n&1\end{array}\right) = n$$
and
$$\left(\begin{array}{c}n&2\end{array}\left) = \frac{(n-1)(n-2)}{2}}$$

7. Jan 11, 2005

### recon

Do you mean that I have to expand $$(1 + \sqrt\frac{2}{n-1})^n$$? It's the square root that is confusing me. I can't get rid of it. :yuck:

8. Jan 11, 2005

### recon

Putting the question in another form, how do I proof that $$\sqrt{\frac{2}{n-1}}$$ decreases in value slower than $${n^{{\frac{1}{n}}}-1$$ as n increases?