11n + 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
prove that for all integers n>1,
1/n+1+1/n+2+/n+3.....1/2n>1/2
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UP 0 DOWN 0 0 12
12 Answers
1
49isn'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!
1" 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 .
1Yup I realized that the proper upper bound would be " ln 2 " , which in fact is attainable .
49yes that is what i ws saying ...
in my post!
wasn't i correct?
66@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.