Expand & Simplify Binomial (1 + $\sqrt\frac{2}{n-1}$)^n

[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?

 

[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.

 

[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 0as n \rightarrow \infty. Is there a more rigorous way for showing this?

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

 

[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.

 

[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}

 

[NateTG]

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}}

 

[recon]

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.

 

[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?