Deja vu?

Algebra Level pending

There is a sequence { b n } \{b_n\} such that b n = 1 n + 1 b_n=\dfrac{1}{n+1} . Let T n T_n be the sum of first n n terms of { b n } \{b_n\} . If there exists positive integer n n such that 1 λ + T n T 2 n \dfrac{1}{\lambda}+T_n \geq T_{2n} for real λ \lambda , find the maximum value of λ \lambda .


The answer is 3.

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1 solution

The quantity 1 λ \dfrac{1}{\lambda} increases with increase in n n . So λ \lambda is maximum when n n is minimum with value 1 1 . The maximum value of λ \lambda is 1 1 2 + 1 = 3 \dfrac{1}{\dfrac{1}{2+1}}=\boxed 3

I disagree, T 2 n T n T_{2n}-T_n is always greater than 1/2, therefore λ \lambda has to be smaller than 2.

Moreover, when n = 1 n=1 , T 2 n T n = 1 2 + 1 1 1 1 = 1 2 T_{2n}-T_n= \frac{1}{2}+\frac{1}{1}-\frac{1}{1}= \frac{1}{2} which gives λ = 2 \lambda=2 not 3.

Théo Leblanc - 1 year, 5 months ago

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Check what you have written. At n = 1 n=1 , T n = T 1 = b 1 = 1 2 T_n=T_1=b_1=\dfrac{1}{2} . T 2 n = T 2 = b 1 + b 2 = 1 2 + 1 3 T_{2n}=T_2=b_1+b_2=\dfrac{1}{2}+\dfrac{1}{3} . So T 2 n T n = 1 3 T_{2n}-T_n=\dfrac{1}{3} , and hence λ = 3 \lambda=3 , and not 2 2 .

A Former Brilliant Member - 1 year, 5 months ago

The question is wrongly posed, we don't know at which n n starts ( b n ) (b_n) , and this is an issue because it determine the answer: starts at n = 0 , λ = 2 n=0, \ \Rightarrow \lambda=2 , at n = 1 , λ = 3 n=1, \ \Rightarrow\lambda=3 etc. Finally in the question λ \lambda is real and only integers can be given as answer (a legitimate guess for the question is 1 ln ( 2 ) \frac{1}{\ln(2)} ).

In retrospect, I understand that ( b n ) (b_n) starts at n = 1 n=1 but this is completely illogical because when we right 1 / ( n + 1 ) 1/(n+1) it is because we want to start at n = 0 n=0 otherwise we write 1 / n 1/n and because it is not defined at n = 0 n=0 , everyone knows it starts at n = 1 n=1 .

The classic " 1 / n " "1/n" sequence (almost) always starts by 1 / 1 1/1 , when it is not the case, it should be clearly written.

Théo Leblanc - 1 year, 5 months ago

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