inequality

prove that for all integers n>1,

1/n+1+1/n+2+/n+3.....1/2n>1/2

12 Answers

1
Ricky ·

1n + 1 > 12 n

1n + 2 > 12n

. . . . . . . . . ( after n - 3 steps )

12 n = 12 n

Adding , we get ,

1n + 1 + 1n + 2 + .............+ 12 n > n2 n = 12

66
kaymant ·

Is there some upper bound?

49
Subhomoy Bakshi ·

isn't it 1/2 + 1/3 + 1/4 +....till n terms?

the sum can be found out as in here!

http://www.targetiit.com/iit-jee-forum/posts/sum-of-series-hav-a-taste-of-this-13231.html

and later using integration with limits 1 to n of the function approximated!

11
Soham Mukherjee ·

how can we get n+2>2n?

1
Ricky ·

" n + 2 " is not greater than " 2 n " . It is infact less than " 2 n " as n > 1 and " n " is an integer . So its RECIPROCAL IS GREATER THAN THE RECIPROCAL OF " 2n " , i . e ,

n + 2 < 2n ..........implies

1n + 2 > 12 n

As to Kaymant Sir 's query , YES , there is an upper bound of 34 .

1
jangra28192manoj jangra ·

thats nice ricky

341
Hari Shankar ·

Sir has in mind a tighter bound of about 0.693

1
Ricky ·

Yup I realized that the proper upper bound would be " ln 2 " , which in fact is attainable .

49
Subhomoy Bakshi ·

yes that is what i ws saying ...

in my post!

wasn't i correct?

66
kaymant ·

@TheProphet,
sir, you are damn good at mind reading [1]

341
Hari Shankar ·

Thats the only bound I knew :D

66
kaymant ·

@Ricky,
By the way, that bound of ln2 is not attainable as you may think but you could reach arbitrarily close by making n arbitrarily large. ln 2 is the limiting value.

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