# Thread: finding the Inequality of arithmetic and geometric means of a sequence

1. ## finding the Inequality of arithmetic and geometric means of a sequence

Hello everyone

I was requested to prove this inequality

for this sequence

using Inequality arithmetic and geometric means (using arithmetic mean and geometric mean only)

I tried to solve this thing out in so many ways I will be really glad if someone will show me the full solution for this Inequality

thanks

2. ## Re: finding the Inequality of arithmetic and geometric means of a sequence

Originally Posted by someone111888
I was requested to prove this inequality

for this sequence
As posted, the statement is not true.

$a_1=\dfrac{1}{2},~a_2=\dfrac{7}{12},~a_3=\dfrac{3 7}{60},\cdots~a_9=\dfrac{1632341}{2450448}$

Could it be that $N\ge 10~?$ Are you sure that you have copied it correctly?

3. ## Re: finding the Inequality of arithmetic and geometric means of a sequence

Yes I did, my intesntion is that there is a requestment to prove that the sequence itself bigger than 10 using this equality

4. ## Re: finding the Inequality of arithmetic and geometric means of a sequence

Im sorry bigger than 2/3

5. ## Re: finding the Inequality of arithmetic and geometric means of a sequence

Originally Posted by someone111888
Yes I did, my intesntion is that there is a requestment to prove that the sequence itself bigger than 10 using this equality
Originally Posted by someone111888
Im sorry bigger than 2/3
Now I am totally confused.

Just post the exact wording of the original question.

6. ## Re: finding the Inequality of arithmetic and geometric means of a sequence

Originally Posted by Plato
Now I am totally confused.
What else is new?

7. ## Re: finding the Inequality of arithmetic and geometric means of a sequence

Originally Posted by someone111888
Im sorry bigger than 2/3
Then DenisB just gave you a proof by contradiction showing that inequality is false for n <10. You cannot prove a false statement to be true. He gave you explicit examples showing that it is false.

$a_{n+1}=a_n+\dfrac{1}{2n+1}+\dfrac{1}{2n+2}-\dfrac{1}{n+1} = a_n+\dfrac{1}{(2n+1)(2n+2)}$

So your sequence is increasing as n increases. You should be able to find the minimum value of n that makes the equality true

http://m.wolframalpha.com/input/?i=l...to+infinity%5D

8. ## Re: finding the Inequality of arithmetic and geometric means of a sequence

Notice first that for any positive integer n, $\int_n^{2n}1/x\,dx=\ln(2)$. Next by consideration of upper and lower Riemann sums for this integral, it is easy to show that
$$\ln(2)-{1\over2n}\leq a_n\leq\ln(2)$$

Hence
$$\lim_{n\to\infty}a_n=\ln(2)$$

Since $\ln(2)\approx0.693$, from the above inequality, it is obvious that eventually $a_n>2/3$. As was noted in above posts, the inequality is true for any $n\geq10$. To me, this was kind of a strange question; I think a better question would be to find the limit of the sequence, as done above.