hkm_gmat wrote:

laxieqv: Please advise us how we can learn such strategies and give us some links to learn them. Thanks in advance !

Let me type the question again first:

1/(n+1)< 1/31+1/32+1/33< 1/n

My reasonsing process is:

+Why don't we compare a sum of

3 numbers with a sum of

3 numbers instead of sth like the question?! -----> I try to express

1/(n+1) and 1/n as sums of three numbers:

1/(n+1) = 3/ [3(n+1)] = 1/(3n+3) + 1/(3n+3) + 1/(3n+3)

In the similar manner, we have 1/n = 1/(3n)+ 1/(3n)+ 1/(3n)

+since 3n+3 and 3n are very very close-----> I try to make the two ends of the inequality as close to ( 1/31+1/32+1/33) as possible -----> of coz, the best way is to make 3n+3= 30 and 3n= sth very close to 31( that is to say 30)---> i got n=10.

I don't know any website which teaches these techniques

. As gamjatang told me that this question is 50-51 range, it's worth some reasoning

But from now on, you can hit similar questions like:

1/(n+1) < 1/41+1/42+1/43+1/44< 1/n

or 1/(n+1) < 1/51+1/52+1/53+1/54+1/55< 1/n

and so forth