MHB How to Prove the Inequality of the Sequence T_n?

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The discussion centers on proving the inequality for the sequence T_n, defined as the sum of terms of the form \(1 - \frac{1}{(2k+1)^2}\) for k from 1 to n. Participants initially question the validity of the right-hand inequality, suggesting it is incorrect since each term in T_n approaches 1. The correct approach involves defining two products, P_n and Q_n, which bound T_n and lead to the conclusion that \( \sqrt{\frac{n+1}{2n+1}} < T_n < \sqrt{\frac{2n+3}{3n+3}} \). The final consensus affirms the inequality is valid, emphasizing the importance of proper definitions in mathematical proofs. The discussion concludes with a note that the proof becomes straightforward once the correct relationships are established.
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$ 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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Albert said:
$ 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 said:
$ 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 said:
$T_n=\left(1-\dfrac{1}{3^2} \right)\times \left(1-\dfrac{1}{5^2} \right)\times \left(1-\dfrac{1}{7^2} \right)\times \cdots\times \left[1-\dfrac{1}{(2n+1)^2} \right]$
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 $P_n<T_n<Q_n$

$\therefore \sqrt{\dfrac {n+1}{2n+1}}<T_n < \sqrt{\dfrac {2n+3}{3n+3}}$
 
Albert said:
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!
 

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