IIT JEE , AIEEE Forum - Mathematics , Algebra

konichiwa2x

Hi, I got this idea from conjurer. You can post any inequality question that you came across. So this can be like a database for inequalities.

Here is one. If

prove that

Decoder

the upper bound..

multiply both sides by 2^ 1998
rhs becomes (odd prod.).2^999 / 999!
this is nothing but 1998C999...

now it becomes something like a = 2nCn / 2^2n.
rhs is always less than 1/2n+1..( 2^2n/2nCn > 2n+1 )
hence it becomes a>1/1999...

hsbhatt

$ext {The 2nd part is easy} \ \ ext {We have} 1 rac{3} {2} . rac {5} {4} ... rac{1999} {1998} ge 1 \ \ ext {Hence} a ge rac {1} {1999}$

Correct >= to >

sandeepramesh

also 2nd part can be done by Wierstrass

hsbhatt

good going madness. post it, many will benefit

sandeepramesh

i dont know how to use latex properly Help?

hsbhatt

use texer and paste. but our editor is good enough for weierstrass.

sboosy

a = 1/2 *3/4 *5/6 * ....1997/1998

so a > 1/2 *3/4 *5/6 * .....1997/1998 * 1998/1999

because multiplying by a number less than 1

now cancel that 1998 with 1998

and shift all terms in numerator by 1 place to left

we get

a> 3/2 * 5/4 * ......1997/1996 * 1/1999

which is > 1*1*1* ....1/1999

i m making all fractions > 1 as 1

so

a>1/1999

now

a = 1/2 *3/4 *5/6 * ....21/22 * ......

a = 1/2 *21/22 * 3/4 * 5/6 * ...... shifting 21/22 left

a < 1/2 *21/22 * 4/3 * 5/6 * 8/7 * 9/10 .....

i m making 3/4 as 4/3 and 7/8 as 8/7

so cancelling 3 and 7 with 21

a< 1/2 * 1/22 * 4* 8 * 5/6 * 9/10 * .....

a< 1/44 *( 4 *8 * 11/12 * 13/14 .1997/1998)...

denominator using product series is more than 32 times the numerator

so the bracket is <1

so a <1/44

sandeepramesh

hsbhatt will post the soln later ok? Am busy now :D

konichiwa2x

that was very nice sboosy!

konichiwa2x

Here is another way of proving it.

If

, then to prove the inequality

it seems natural to guess that

It should be easy enough to prove that by induction.

The other inequality,

, looks harder. One approach would be to write

. Then you can use the inequality

. That would give

Hence you can go on to show that

Edit: a better approach is to notice that ?(1998)=44.69... . That suggests the inequality

another straightforward induction proof

elastiboysai

Hi

I have a proof widout induction.

Ive encountered dis method b4 bt

culdn rem xactly...

elastiboysai

Now use the derived inequality..

u get the reqd result

If im rt

u get sumthin like

1/2

1000<S<1/

1999

frm dat the reqd result is obv.

edit

continuation of proof attached

elastiboysai

But I guess

Wiestrass inequality is the most elegant way of proving the second part

Wiestrass inequality states

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