# Prove inequality of T_n

#### Albert

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

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

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#### Opalg

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

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 $1$).

#### Albert

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

prove: $\sqrt{\dfrac{n+1}{2n+1}}<T_n<\sqrt{\dfrac{2n+3}{3n+3}}$
sory : a typo !

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

#### Albert

##### Well-known member
$T_n=\left(1-\dfrac{1}{3^2} \right)\times \left(1-\dfrac{1}{5^2} \right)\times \left(1-\dfrac{1}{7^2} \right)\times \cdots\times \left[1-\dfrac{1}{(2n+1)^2} \right]$
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},\,\,\, $$T_nQ_n=\dfrac{2n+3}{3n+3} but P_n<T_n<Q_n \therefore \sqrt{\dfrac {n+1}{2n+1}}<T_n < \sqrt{\dfrac {2n+3}{3n+3}} #### Opalg ##### MHB Oldtimer Staff member prove : \sqrt{\dfrac {n+1}{2n+1}}<T_n < \sqrt{\dfrac {2n+3}{3n+3}} let P_n=\left(1-\dfrac{1}{2^2} \right)\times \left(1-\dfrac{1}{4^2} \right)\times \left(1-\dfrac{1}{6^2} \right)\times \cdots\times \left[1-\dfrac{1}{(2n)^2} \right] let Q_n=\left(1-\dfrac{1}{4^2} \right)\times \left(1-\dfrac{1}{6^2} \right)\times \left(1-\dfrac{1}{8^2} \right)\times \cdots\times \left[1-\dfrac{1}{(2n+2)^2} \right] T_nP_n=\dfrac{n+1}{2n+1},\,\,\,$$ T_nQ_n=\dfrac{2n+3}{3n+3}$

but $P_n<T_n<Q_n$

$\therefore \sqrt{\dfrac {n+1}{2n+1}}<T_n < \sqrt{\dfrac {2n+3}{3n+3}}$
Easy, once you have seen it!