How to Prove the Inequality of the Sequence T_n?

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Discussion Overview

The discussion revolves around proving the inequality of the sequence \( T_n \), defined as a sum involving terms of the form \( 1 - \frac{1}{(2k+1)^2} \) for \( k = 1 \) to \( n \). Participants are tasked with demonstrating that \( \sqrt{\frac{n+1}{2n+1}} < T_n < \sqrt{\frac{2n+3}{3n+3}} \), exploring both the formulation of \( T_n \) and the validity of the proposed inequalities.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants present the definition of \( T_n \) as a sum of terms \( 1 - \frac{1}{(2k+1)^2} \) and propose the inequalities to be proven.
  • One participant questions the validity of the right-hand inequality, arguing that since each term in the sum is close to 1, the total \( T_n \) should not be less than \( \sqrt{\frac{2n+3}{3n+3}} \), which is suggested to be less than 1.
  • Another participant corrects a typographical error in the definition of \( T_n \), initially misrepresented as a product rather than a sum.
  • Further contributions reiterate the inequalities and introduce new sequences \( P_n \) and \( Q_n \), which are products of similar terms, to establish bounds for \( T_n \). They assert that \( T_n P_n = \frac{n+1}{2n+1} \) and \( T_n Q_n = \frac{2n+3}{3n+3} \), leading to the inequalities being restated.
  • Some participants express confidence in the inequalities, suggesting that they become clear once the reasoning is understood.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the validity of the right-hand inequality, with at least one participant challenging its correctness. The discussion includes multiple perspectives on the formulation and proof of the inequalities.

Contextual Notes

There are unresolved issues regarding the assumptions made about the terms in the sequence and the implications of the inequalities, particularly concerning the behavior of \( T_n \) as \( n \) increases.

Albert1
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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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