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Math Help - Mathematical Induction

  1. #1
    Junior Member
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    Mathematical Induction

    I need a little help on this problem:

    1/2 x 3/4 x ... x (2n-1)/2n <= 1/(sqrt(3n+1)

    I know that I have to show that for n= 1 works first and it does. Then I know that I need to show for n+1.

    This is what I did, I first added n+1 to both sides

    this is what I got: 1/2 x 3/4 x ... x (2n-1)/2n x [2(n+1) - 1]/2(n+1)<=
    1/(sqrt(3(n+1)+1))

    now i know that the first part of what I did on the LHS is <= to the RHS, so I substituted: and got this:

    1/(sqrt(3n+1) x (2n+1)/(2n+2) <= 1(sqrt(3n+4))

    Am I on the right track here?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Nichelle14
    I need a little help on this problem:

    1/2 x 3/4 x ... x (2n-1)/2n <= 1/(sqrt(3n+1)

    I know that I have to show that for n= 1 works first and it does. Then I know that I need to show for n+1.

    This is what I did, I first added n+1 to both sides

    this is what I got: 1/2 x 3/4 x ... x (2n-1)/2n x [2(n+1) - 1]/2(n+1)<=
    1/(sqrt(3(n+1)+1))

    now i know that the first part of what I did on the LHS is <= to the RHS, so I substituted: and got this:

    1/(sqrt(3n+1) x (2n+1)/(2n+2) <= 1(sqrt(3n+4))

    Am I on the right track here?
    You have done the base case so I won't worry about that, so let's assume
    this is true for some k \ge 1, that is for this k:

    <br />
\frac{1}{2}\times \frac{3}{4}\times \dots \times \frac{2k-1}{2k} \le \frac{1}{\sqrt{3k+1}}\ \ \ \dots(1)<br />
.

    Now multiply both sides by (2k+1)/(2k+2), which would be the next term
    of the product on the Left Hand Side (LHS) to give:

    <br />
\frac{1}{2}\times \frac{3}{4}\times \dots \times \frac{2k-1}{2k} \times \frac{2k+1}{2k+2} \le \frac{1}{\sqrt{3k+1}}\times \frac{2k+1}{2k+2}\ \ \ \dots(2)<br />
.

    Now if we can prove that the RHS of (2) is less than or equal \frac{1}{\sqrt{3k+2}}
    this will be sufficient to prove:

    <br />
\frac{1}{2}\times \frac{3}{4}\times \dots \times \frac{2k+1}{2k+2} \le \frac{1}{\sqrt{3k+4}}\ \ \ \dots(3)<br />
,

    which is what we need for the induction step.

    So we now seek to proove:

    <br />
\frac{1}{\sqrt{3k+1}}\times \frac{2k+1}{2k+2} \le \frac{1}{\sqrt{3k+2}}\ \ \ \dots(4)<br />

    Now (4) is true iff (if and only if):

    <br />
\frac{2k+1}{2k+2} \le \left( \frac{3k+1}{3k+2} \right)^2<br />
,

    which itself is true iff:

    <br />
\left( \frac{2k+1}{2k+2} \right)^2 \le  \frac{3k+1}{3k+2}<br />
.

    Which is true iff:

    (2k+1)^2(3k+2) \le (2k+2)^2(3k+1)

    multiplying both sides of the above out give this is true iff:

    <br />
12k^3+20k^2+11k+2 \le 12k^2+28k^2+20k+4<br />

    which is true as the coefficients of the powers of k on the LHS
    are all \le the corresponding coefficients on the RHS, and so
    the induction step is proven.

    RonL
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