Deja vu?

Algebra Level pending

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

The answer is 3.

1 solution

A Former Brilliant Member
Jan 9, 2020

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

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

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

Théo Leblanc - 1 year, 5 months ago

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

A Former Brilliant Member - 1 year, 5 months ago

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 $n=0, \ \Rightarrow \lambda=2$ , at $n=1, \ \Rightarrow\lambda=3$ etc. Finally in the question $\lambda$ is real and only integers can be given as answer (a legitimate guess for the question is $\frac{1}{\ln(2)}$ ).

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.

Théo Leblanc - 1 year, 5 months ago

Deja vu?

The answer is 3.

1 solution

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