## binomial expansion

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?

 Blog Entries: 9 Recognitions: Homework Help Science Advisor 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.
 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 $$n^\frac{1}{n}$$?

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## binomial expansion

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.

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

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

 Is this problem solvable? If n is a positive integer, prove that: $$n^\frac{1}{n} < 1 + \sqrt\frac{2}{n-1}$$
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
$$\left(\begin{array}{c}n&2\end{array}\left) = \frac{(n-1)(n-2)}{2}}$$

 Quote by NateTG 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 $$\left(\begin{array}{c}n&2\end{array}\left) = \frac{(n-1)(n-2)}{2}}$$
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.

 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?