There is a sequence { b n } such that b n = n + 1 1 . Let T n be the sum of first n terms of { b n } . If there exists positive integer n such that λ 1 + T n ≥ T 2 n for real λ , find the maximum value of λ .
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I disagree, T 2 n − T n is always greater than 1/2, therefore λ has to be smaller than 2.
Moreover, when n = 1 , T 2 n − T n = 2 1 + 1 1 − 1 1 = 2 1 which gives λ = 2 not 3.
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Check what you have written. At n = 1 , T n = T 1 = b 1 = 2 1 . T 2 n = T 2 = b 1 + b 2 = 2 1 + 3 1 . So T 2 n − T n = 3 1 , and hence λ = 3 , and not 2 .
The question is wrongly posed, we don't know at which n starts ( b n ) , and this is an issue because it determine the answer: starts at n = 0 , ⇒ λ = 2 , at n = 1 , ⇒ λ = 3 etc. Finally in the question λ is real and only integers can be given as answer (a legitimate guess for the question is ln ( 2 ) 1 ).
In retrospect, I understand that ( b n ) starts at n = 1 but this is completely illogical because when we right 1 / ( n + 1 ) it is because we want to start at n = 0 otherwise we write 1 / n and because it is not defined at n = 0 , everyone knows it starts at n = 1 .
The classic " 1 / n " sequence (almost) always starts by 1 / 1 , when it is not the case, it should be clearly written.
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The quantity λ 1 increases with increase in n . So λ is maximum when n is minimum with value 1 . The maximum value of λ is 2 + 1 1 1 = 3