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Binomial expansion

  1. Jan 10, 2005 #1
    How do you expand and simplify [tex](1 + \sqrt\frac{2}{n-1})^n[/tex]?

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

    [tex]\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} + ...[/tex]

    but how do you simplify this?
    Last edited: Jan 10, 2005
  2. jcsd
  3. Jan 10, 2005 #2


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    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.

  4. Jan 11, 2005 #3
    Hmm. How can I solve the following problem then?

    n is a positive integer. Prove that:

    [tex]n^\frac{1}{n} < 1 + \sqrt\frac{2}{n-1}[/tex]

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

    Also, what is the maximum value of [tex]n^\frac{1}{n} [/tex]?
  5. Jan 11, 2005 #4


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    Do you know calculus??If u did,then
    [tex] \lim_{n\rightarrow +\infty} n^{\frac{1}{n}}=\alpha [/tex](1)
    U need to show that \alpha=1.
    Take natural logarithm from both sides.Then
    [tex] \lim_{n\rightarrow +\infty} \frac{1}{n}\ln n =\ln\alpha [/tex] (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 [itex]\ln\alpha=0 \Rightarrow \alpha=1[/itex].

  6. Jan 11, 2005 #5
    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:
    [tex]n^\frac{1}{n} < 1 + \sqrt\frac{2}{n-1}[/tex]
  7. Jan 11, 2005 #6


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    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:
    [tex]\left(\begin{array}{c}a&b\end{array}\right) = \frac{a!}{b!(a-b)!}[/tex]
    [tex]\left(\begin{array}{c}n&0\end{array}\right) = 1[/tex]
    [tex]\left(\begin{array}{c}n&1\end{array}\right) = n[/tex]
    [tex]\left(\begin{array}{c}n&2\end{array}\left) = \frac{(n-1)(n-2)}{2}}[/tex]
  8. Jan 11, 2005 #7
    Do you mean that I have to expand [tex](1 + \sqrt\frac{2}{n-1})^n[/tex]? It's the square root that is confusing me. I can't get rid of it. :yuck:
  9. Jan 11, 2005 #8
    Putting the question in another form, how do I proof that [tex]\sqrt{\frac{2}{n-1}}[/tex] decreases in value slower than [tex]{n^{{\frac{1}{n}}}-1[/tex] as n increases?
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