# Thread: Mathematical Induction

1. ## 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?

2. 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 $\displaystyle k \ge 1$, that is for this $\displaystyle k$:

$\displaystyle \frac{1}{2}\times \frac{3}{4}\times \dots \times \frac{2k-1}{2k}$$\displaystyle \le \frac{1}{\sqrt{3k+1}}\ \ \ \dots(1) . Now multiply both sides by \displaystyle (2k+1)/(2k+2), which would be the next term of the product on the Left Hand Side (LHS) to give: \displaystyle \frac{1}{2}\times \frac{3}{4}\times \dots \times \frac{2k-1}{2k} \times \frac{2k+1}{2k+2}$$\displaystyle \le \frac{1}{\sqrt{3k+1}}\times \frac{2k+1}{2k+2}\ \ \ \dots(2)$.

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

$\displaystyle \frac{1}{2}\times \frac{3}{4}\times \dots \times \frac{2k+1}{2k+2}$$\displaystyle \le \frac{1}{\sqrt{3k+4}}\ \ \ \dots(3)$,

which is what we need for the induction step.

So we now seek to proove:

$\displaystyle \frac{1}{\sqrt{3k+1}}\times \frac{2k+1}{2k+2} \le \frac{1}{\sqrt{3k+2}}\ \ \ \dots(4)$

Now $\displaystyle (4)$ is true iff (if and only if):

$\displaystyle \frac{2k+1}{2k+2} \le \left( \frac{3k+1}{3k+2} \right)^2$,

which itself is true iff:

$\displaystyle \left( \frac{2k+1}{2k+2} \right)^2 \le \frac{3k+1}{3k+2}$.

Which is true iff:

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

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

$\displaystyle 12k^3+20k^2+11k+2 \le 12k^2+28k^2+20k+4$

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

RonL

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### 2n-1/2n < 1/sqrt(2n 1)

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