Proving I_n \leq (2n/(2n+1))^n by induction for positive n

[converting1]

I_n = \displaystyle \int_0^1 (1-x^2)^ndx, n \geq 0

Given that (2n + 1)I_n = 2nI_{n-1}

proove by induction that

I_n \leq \left (\dfrac{2n}{2n + 1} \right)^n for positive integers of n

in the solutions, could someone explain how they got to step 1, and why we need to show step 2 to complete the proof?

Solutions: http://gyazo.com/26e85134e4d5c13d5d7a49a0de91ae58

 

[milesyoung]

You have the statement:
<br /> S_n : I_n \leq \left( \frac{2n}{2n + 1} \right)^n<br />
and you want to prove that it's true for all n, where n is a positive integer.
You can do this by proving that S_1 and S_k \Rightarrow S_{k + 1} are true, where k is a positive integer.

Proving S_1 is true should be simple enough. Prove S_k \Rightarrow S_{k + 1} is true by direct proof, i.e. assume S_k to be true and show that it forces S_{k + 1} to be true:
<br /> S_k : I_k \leq \left( \frac{2k}{2k + 1} \right)^k \Leftrightarrow \frac{2k + 2}{2k + 3} I_k \leq \frac{2k + 2}{2k + 3} \left( \frac{2k}{2k + 1} \right)^k \Leftrightarrow I_{k + 1} \leq \frac{2k + 2}{2k + 3} \left( \frac{2k}{2k + 1} \right)^k<br />
Here I just multiplied both sides of the inequality by \frac{2k + 2}{2k + 3} and used the relation I_n = \frac{2n}{2n + 1} I_{n - 1}.

You have:
<br /> S_{k + 1} : I_{k + 1} \leq \left( \frac{2(k + 1)}{2(k + 1) + 1} \right)^{k + 1} = \left( \frac{2k + 2}{2k + 3} \right)^{k + 1} = \left( \frac{2k + 2}{2k + 3} \right)^{1} \left( \frac{2k + 2}{2k + 3} \right)^{k}<br />
Thus, if you can show that:
<br /> \frac{2k + 2}{2k + 3} \left( \frac{2k}{2k + 1} \right)^k \leq \left( \frac{2k + 2}{2k + 3} \right)^{1} \left( \frac{2k + 2}{2k + 3} \right)^{k}<br />
then S_{k + 1} must be true if S_{k} is true.

It follows by induction that S_n is true for all n, where n is a positive integer.

 

[converting1]

You have the statement:
<br /> S_n : I_n \leq \left( \frac{2n}{2n + 1} \right)^n<br />
and you want to prove that it's true for all n, where n is a positive integer.
You can do this by proving that S_1 and S_k \Rightarrow S_{k + 1} are true, where k is a positive integer.

Proving S_1 is true should be simple enough. Prove S_k \Rightarrow S_{k + 1} is true by direct proof, i.e. assume S_k to be true and show that it forces S_{k + 1} to be true:
<br /> S_k : I_k \leq \left( \frac{2k}{2k + 1} \right)^k \Leftrightarrow \frac{2k + 2}{2k + 3} I_k \leq \frac{2k + 2}{2k + 3} \left( \frac{2k}{2k + 1} \right)^k \Leftrightarrow I_{k + 1} \leq \frac{2k + 2}{2k + 3} \left( \frac{2k}{2k + 1} \right)^k<br />
Here I just multiplied both sides of the inequality by \frac{2k + 2}{2k + 3} and used the relation I_n = \frac{2n}{2n + 1} I_{n - 1}.

You have:
<br /> S_{k + 1} : I_{k + 1} \leq \left( \frac{2(k + 1)}{2(k + 1) + 1} \right)^{k + 1} = \left( \frac{2k + 2}{2k + 3} \right)^{k + 1} = \left( \frac{2k + 2}{2k + 3} \right)^{1} \left( \frac{2k + 2}{2k + 3} \right)^{k}<br />
Thus, if you can show that:
<br /> \frac{2k + 2}{2k + 3} \left( \frac{2k}{2k + 1} \right)^k \leq \left( \frac{2k + 2}{2k + 3} \right)^{1} \left( \frac{2k + 2}{2k + 3} \right)^{k}<br />
then S_{k + 1} must be true if S_{k} is true.

It follows by induction that S_n is true for all n, where n is a positive integer.

managed to show that's true

thank you :)