MHB How to Prove the Inequality of the Sequence T_n?

[Albert1]

$ T_n=\left(1-\dfrac{1}{3^2} \right)+\left(1-\dfrac{1}{5^2} \right)+\left(1-\dfrac{1}{7^2} \right)+\cdots+\left[1-\dfrac{1}{(2n+1)^2} \right]$

prove: $ \sqrt{\dfrac{n+1}{2n+1}}<T_n<\sqrt{\dfrac{2n+3}{3n+3}}$

 

[Opalg]

$ T_n=\left(1-\dfrac{1}{3^2} \right)+\left(1-\dfrac{1}{5^2} \right)+\left(1-\dfrac{1}{7^2} \right)+\cdots+\left[1-\dfrac{1}{(2n+1)^2} \right]$

prove: $ \sqrt{\dfrac{n+1}{2n+1}}<T_n<\sqrt{\dfrac{2n+3}{3n+3}}$

Could you please check the wording of this problem. The right-hand inequality is clearly false (each of the $n$ terms in the sum $T_n$ is close to $1$, yet their sum is supposed to be less than $\sqrt{(2n+3)/(3n+3)}$, which is less than $1$).

 

[Albert1]

$ T_n=\left(1-\dfrac{1}{3^2} \right)+\left(1-\dfrac{1}{5^2} \right)+\left(1-\dfrac{1}{7^2} \right)+\cdots+\left[1-\dfrac{1}{(2n+1)^2} \right]$

prove: $ \sqrt{\dfrac{n+1}{2n+1}}<T_n<\sqrt{\dfrac{2n+3}{3n+3}}$

sory : a typo !

$T_n=\left(1-\dfrac{1}{3^2} \right)\times \left(1-\dfrac{1}{5^2} \right)\times \left(1-\dfrac{1}{7^2} \right)\times \cdots\times \left[1-\dfrac{1}{(2n+1)^2} \right]$

 

[Albert1]

$T_n=\left(1-\dfrac{1}{3^2} \right)\times \left(1-\dfrac{1}{5^2} \right)\times \left(1-\dfrac{1}{7^2} \right)\times \cdots\times \left[1-\dfrac{1}{(2n+1)^2} \right]$

prove :
$ \sqrt{\dfrac {n+1}{2n+1}}<T_n < \sqrt{\dfrac {2n+3}{3n+3}}$

let $P_n=\left(1-\dfrac{1}{2^2} \right)\times \left(1-\dfrac{1}{4^2} \right)\times \left(1-\dfrac{1}{6^2} \right)\times \cdots\times \left[1-\dfrac{1}{(2n)^2} \right]$

let $Q_n=\left(1-\dfrac{1}{4^2} \right)\times \left(1-\dfrac{1}{6^2} \right)\times \left(1-\dfrac{1}{8^2} \right)\times \cdots\times \left[1-\dfrac{1}{(2n+2)^2} \right]$

$ T_nP_n=\dfrac{n+1}{2n+1},\,\,\, $$ T_nQ_n=\dfrac{2n+3}{3n+3}$

but $P_n<T_n<Q_n$

$\therefore \sqrt{\dfrac {n+1}{2n+1}}<T_n < \sqrt{\dfrac {2n+3}{3n+3}}$

 

[Opalg]

prove :
$ \sqrt{\dfrac {n+1}{2n+1}}<T_n < \sqrt{\dfrac {2n+3}{3n+3}}$

let $P_n=\left(1-\dfrac{1}{2^2} \right)\times \left(1-\dfrac{1}{4^2} \right)\times \left(1-\dfrac{1}{6^2} \right)\times \cdots\times \left[1-\dfrac{1}{(2n)^2} \right]$

let $Q_n=\left(1-\dfrac{1}{4^2} \right)\times \left(1-\dfrac{1}{6^2} \right)\times \left(1-\dfrac{1}{8^2} \right)\times \cdots\times \left[1-\dfrac{1}{(2n+2)^2} \right]$

$ T_nP_n=\dfrac{n+1}{2n+1},\,\,\, $$ T_nQ_n=\dfrac{2n+3}{3n+3}$

but $P_n<T_n<Q_n$

$\therefore \sqrt{\dfrac {n+1}{2n+1}}<T_n < \sqrt{\dfrac {2n+3}{3n+3}}$

View attachment 707 Easy, once you have seen it!