IIT JEE 2008 P1 Q9

1 Answers

For n=1, we get,

S1=1/(1+1+1)
=0.3

and T1=1/(1+0)=1

Also, pi/3√3=pi√3/9=0.58

So,S1<pi/√3*3<T1

Hence,Sn<pi/3*√3 and T1>pi/√3*3

I know it's not the correct method..but since noone solved it, dats why....!!

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