Home » Ask & Discuss » Mathematics. » Algebra « Back to Discussion

Algebra

Blazing goIITian

Joined:	12 Apr 2008
Post:	2717

30 Oct 2008 18:33:28 IST

0 People liked this

1054

Inequality

None

Prove that:

$\frac{\sqrt{1}+\sqrt{\frac{1}{2}}+\cdots\cdots+\sqrt{\frac{1}{n}}}{\sqrt{n}}<(2n-1)^{1/4}$

and of course n is a +ve number. :D

Share this article on:

Comments (12)

Hari Shankar

Forum Expert

Joined: 28 Feb 2007

Posts: 2173

31 Oct 2008 09:23:13 IST

Like 4 people liked this

First we will prove the inequality $1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n} < \sqrt{2n-1}$

$n+1 = \sqrt{(n+1)^2} = \sqrt{n^2+2n+1} > \sqrt{2n+1} > \sqrt{2n-1}$

$\therefore 2(n+1)> \sqrt{2n+1} + \sqrt{2n-1}$

and from this we get

$\frac{2}{\sqrt{2n+1} + \sqrt{2n-1}}>\frac{1}{n+1} \\ \\ \text{or} \ \sqrt{2n+1} - \sqrt{2n-1} > \frac{1}{n+1} $

This can be used as the induction step or we can just add the inequalities:

$\frac{1}{n} < \sqrt{2n-1}-\sqrt{2n-3} \\ \\ \frac{1}{n-1} < \sqrt{2n-3} -\sqrt{2n-5} \\ \\ . \\ \\ . \\ \\ . \\ \\ \frac{1}{2} < \sqrt 3 - 1 1 = 1 $

and arrive at $1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n} < \sqrt{2n-1}$

Next from Cauchy Schwarz, we get

$\underbrace{(1+1+....+1)}_{n 1$

or $\sqrt{n} \sqrt{\left(1+\frac{1}{2}+...+\frac{1}{n} \right)} > 1+\frac{1}{\sqrt 2}+...+\frac{1}{\sqrt n}$

Combining this with the previous inequality, we get

$\sqrt{n} \ \sqrt [4] {2n-1}> 1+\frac{1}{\sqrt 2}+...+\frac{1}{\sqrt n}$ or

$\frac{1+\frac{1}{\sqrt 2}+...+\frac{1}{\sqrt n}}{\sqrt n} <\sqrt [4] {2n-1}$

Soumik

Blazing goIITian

Joined: 31 Jul 2008

Posts: 1266

31 Oct 2008 15:01:07 IST

Like 0 people liked this

Well, I've a proof that uses Tchebycheff's inequality, but posting it would be mere foolery seeing Bhatt sir has given the soln....

Anant Kumar

Forum Expert

Joined: 10 Jul 2008

Posts: 598

31 Oct 2008 15:11:12 IST

Like 0 people liked this

Why don't you give your solution.. Though Bhatt sir has given a solution, we still may be enlightened by your solution.

Soumik

Blazing goIITian

Joined: 31 Jul 2008

Posts: 1266

31 Oct 2008 15:58:20 IST

Like 8 people liked this

$\sqrt{1}>\sqrt{\frac{1}{2}}>....\sqrt{\frac{1}{n}} \text{For 2 sets we have}\sqrt{1},\sqrt{\frac{1}{2}}....\sqrt{\frac{1}{n}};\sqrt{1},\sqrt{\frac{1}{n}}\\ \implies\ (\sqrt{1}+\sqrt{\frac{1}{2}}+...+\sqrt{\frac{1}{n}})^2<n(1+\frac{1}{2}+....\frac{1}{n})$ ............(1)

Again applying Tchebychef's inequality, $\ (1+\frac{1}{2}+....\frac{1}{n})^2<n(1.1+\frac{1}{2}.\frac{1}{2}+....\frac{1}{n^2})\\ \text{Also} -n<0\implies\frac{1}{n(n-1)}<\frac{1}{n^2}$ .......(2)

Putting n=2,3,4.....n,

$\frac{1}{1.2}>\frac{1}{2^2}\\ \frac{1}{2.3}>\frac{1}{3^2}\\ \ .....\\ \.....\\ \frac{1}{(n-1)n}>\frac{1}{n^2}\\ \implies$

Adding all corresponding sides, we have...

$\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{(n-1)n}>\frac{1}{2^2}+\frac{1}{3^2}+...\frac{1}{n^2}\\ \implies\ 1-\frac{1}{2}+\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{n-1}-\frac{1}{n}>\frac{1}{2^2}+\frac{1}{3^2}+...\frac{1}{n^2}$

$\ n(1+\frac{1}{2^2}+...\frac{1}{n^2})<(2n-1).........(3)$

From (2) and (3), $\ (1+\frac{1}{2}+....\frac{1}{n})^2<(2n-1)\\ \implies\ (1+\frac{1}{2}+...\frac{1}{n})<(2n-1)^{\frac{1}{2}}$

From (1) &(4), $\ (\sqrt{1}+\sqrt{\frac{1}{2}}+...\sqrt{\frac{1}{n}})^2<n(2n-1)^{\frac{1}{2}}$

From the above result, we get the expected result....

PLZ RATE!

Hari Shankar

Forum Expert

Joined: 28 Feb 2007

Posts: 2173

31 Oct 2008 19:46:16 IST

Like 0 people liked this

hamba , u really underestimated urself. thats a very nice solution.

sourabh

Cool goIITian

Joined: 23 Aug 2008

Posts: 61

5 Nov 2008 15:12:43 IST

Like 0 people liked this

@ Bhat sir,

Can u plz explain hamba's soln a bit in detail? I bet he has jumped a few steps....

sourabh

Cool goIITian

Joined: 23 Aug 2008

Posts: 61

17 Nov 2008 15:06:03 IST

Like 0 people liked this

bhatt, can u understand english?

sourabh

Cool goIITian

Joined: 23 Aug 2008

Posts: 61

17 Dec 2008 14:45:52 IST

Like 0 people liked this

AS I SAID, XPERTS HAVE REALLY NO OTHER USEFUL WORK OTHER THAN GOIN ON RATING THE SAME STUPID ANSWER FROM A STUPID GOIITIAN........IDIOTIC!

Hari Shankar

Forum Expert

Joined: 28 Feb 2007

Posts: 2173

17 Dec 2008 15:27:23 IST

Like 1 people liked this

I dont understand the need for you to go ballistic this way. I didnt reply because

1. I understand that hamba is your next door neighbour so I thought you could just as well ask him

2. There is nothing wrong with his solution

3. Both our solutions are elaborate enough.

and last but not least

4. Your post seemed driven more by a need to put down your friend than to learn something. And we adults are not here to dance according to some adolescent's tunes.

Soumik

Blazing goIITian

Joined: 31 Jul 2008

Posts: 1266

17 Dec 2008 22:54:11 IST

Like 0 people liked this

Thanx Bhatt sir, ur reply suits well the purpose of keeping "The greatest idiot of goiit" QUIET........

Debotosh Chatterjee

Blazing goIITian

Joined: 11 Sep 2008

Posts: 1159

18 Dec 2008 09:26:07 IST

Like 0 people liked this

afterall "you can find your own reason when you want to harm others"----------such guys should be out of this site!!!!!!!!!!!!!

left fr ever!

Scorching goIITian

Joined: 16 Jan 2009

Posts: 287

2 Feb 2009 18:31:12 IST

Like 0 people liked this

yess souvik is debotosh

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team