Help for a really difficult problem

A

Aqua1024

New member
Joined
Jun 6, 2021
Messages
5
20210529_203509~2.jpg20210531_214203~2.jpg
Top image is the problem and the bottom one is my attempt. I tried to prove that Cn+1>Cn, and finally got the inequality. I used mathematical induction to prove but fail. I have no idea for that anymore and keep for a long time. please help me, thanks in advance
 
Aqua1024 said:
View attachment 27650View attachment 27651
Top image is the problem and the bottom one is my attempt. I tried to prove that Cn+1>Cn, and finally got the inequality. I used mathematical induction to prove but fail. I have no idea for that anymore and keep for a long time. please help me, thanks in advance
Click to expand...
Do you realize that \(\displaystyle\mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{1}{n}} \right)^{n}}=\mathop{\lim }\limits_{n \to \infty } {\left( {1 + \frac{1}{n}} \right)^{n + 1}}=e~?\)​
 
[MATH]A_q = \left ( 1 + \dfrac{1}{q} \right )^q = \dfrac{(q + 1)^q}{q^q} \implies A_{q+1} = \dfrac{(q + 2)^{(q+1)}}{(q + 1)^{(q+1)}} .[/MATH]
[MATH]B_q = \left ( 1 + \dfrac{1}{q} \right )^{(q+1)} = \dfrac{(q + 1)^{(q+1)}}{q^{(q+1)}} \implies B_{q+1} = \dfrac{(q + 2)^{(q + 2)}}{(q + 1)^{(q+2)}}.[/MATH]
[MATH]A_q + B_q = \dfrac{(q+1)^q}{q^q} + \dfrac{(q+1)^{(q+ 1)}}{q^{(q+1)}} = \dfrac{q(q + 1)^q + (q+1)^{(q+1)}}{q^{(q+1)}} = \dfrac{(q + 1)^q(q + q + 1)}{q^{(q+1)}} = \dfrac{(q + 1)^q(2q + 2)}{q^{(q+1)}}. [/MATH]
[MATH]2A_qB_q = 2 * \dfrac{(q+1)^q}{q^q} * \dfrac{(q+1)^{(q+ 1)}}{q^{(q+1)}} = \dfrac{2(q + 1)^{(2q+1)}}{q^{(2q+1)}}. [/MATH]
[MATH]\therefore C_q = \dfrac{2(q + 1)^{(2q+1)}}{q^{(2q+1)}} \div \dfrac{(q+1)^q(2q + 2)}{q^{(q+1)}} = WHAT?[/MATH]
[MATH]\therefore C_{q+1} = WHAT?[/MATH]
What then can you deduce about [MATH]\dfrac{C_{q+1}}{C_q}.[/MATH]
 
pka said:
Do you realize that \(\displaystyle\mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{1}{n}} \right)^{n}}=\mathop{\lim }\limits_{n \to \infty } {\left( {1 + \frac{1}{n}} \right)^{n + 1}}=e~?\)​
Click to expand...
Yes, but it's not helpful for me to prove the problem. I also found that An is an increasing sequence whereas Bn is surprisingly a decreasing one. Cn is their harmonic mean, maybe it means something but I didn't realize because I am just a calculus learning beginner, please teach me.
 
JeffM said:
[MATH]A_q = \left ( 1 + \dfrac{1}{q} \right )^q = \dfrac{(q + 1)^q}{q^q} \implies A_{q+1} = \dfrac{(q + 2)^{(q+1)}}{(q + 1)^{(q+1)}} .[/MATH]
[MATH]B_q = \left ( 1 + \dfrac{1}{q} \right )^{(q+1)} = \dfrac{(q + 1)^{(q+1)}}{q^{(q+1)}} \implies B_{q+1} = \dfrac{(q + 2)^{(q + 2)}}{(q + 1)^{(q+2)}}.[/MATH]
[MATH]A_q + B_q = \dfrac{(q+1)^q}{q^q} + \dfrac{(q+1)^{(q+ 1)}}{q^{(q+1)}} = \dfrac{q(q + 1)^q + (q+1)^{(q+1)}}{q^{(q+1)}} = \dfrac{(q + 1)^q(q + q + 1)}{q^{(q+1)}} = \dfrac{(q + 1)^q(2q + 2)}{q^{(q+1)}}. [/MATH]
[MATH]2A_qB_q = 2 * \dfrac{(q+1)^q}{q^q} * \dfrac{(q+1)^{(q+ 1)}}{q^{(q+1)}} = \dfrac{2(q + 1)^{(2q+1)}}{q^{(2q+1)}}. [/MATH]
[MATH]\therefore C_q = \dfrac{2(q + 1)^{(2q+1)}}{q^{(2q+1)}} \div \dfrac{(q+1)^q(2q + 2)}{q^{(q+1)}} = WHAT?[/MATH]
[MATH]\therefore C_{q+1} = WHAT?[/MATH]
What then can you deduce about [MATH]\dfrac{C_{q+1}}{C_q}.[/MATH]
Click to expand...
I tried this way and then got an inequality as well, but I still can't prove the inequality, please teach me.
 
[MATH]C_n[/MATH] is the harmonic mean of [MATH]A_n[/MATH] and [MATH]B_n[/MATH][MATH]C_n=\frac{2}{\tfrac{1}{A_n}+\tfrac{1}{B_n}}[/MATH]Since [MATH]A_n[/MATH] and [MATH]B_n[/MATH] are both positive,
to show that [MATH]C_n[/MATH] is strictly increasing, it is sufficient to show that [MATH]\tfrac{1}{A_n}+\tfrac{1}{B_n}[/MATH] is strictly decreasing
i.e. show that [MATH]\left(\tfrac{n}{n+1} \right)^n + \left(\tfrac{n}{n+1} \right)^{n+1}[/MATH] is strictly decreasing.
 
Last edited:
lex said:
[MATH]C_n[/MATH] is the harmonic mean of [MATH]A_n[/MATH] and [MATH]B_n[/MATH][MATH]C_n=\frac{2}{\tfrac{1}{A_n}+\tfrac{1}{B_n}}[/MATH]Since [MATH]A_n[/MATH] and [MATH]B_n[/MATH] are both positive,
to show that [MATH]C_n[/MATH] is strictly increasing, it is sufficient to show that [MATH]\tfrac{1}{A_n}+\tfrac{1}{B_n}[/MATH] is strictly decreasing
i.e. show that [MATH]\left(\tfrac{n}{n+1} \right)^n + \left(\tfrac{n}{n+1} \right)^{n+1}[/MATH] is strictly decreasing.
Click to expand...
I'll try, thanks a lot!!
 
Aqua1024 said:
Yes, but it's not helpful for me to prove the problem. I also found that An is an increasing sequence whereas Bn is surprisingly a decreasing one. Cn is their harmonic mean, maybe it means something but I didn't realize because I am just a calculus learning beginner, please teach me.
Click to expand...
Well frankly I had expected that you should discover these on your own.
Look at this plot. From which we see that \((C_n)\) is increasing and bounded above. Thus it converges.
The \((A_n)\) sequence is one of the most important in all of mathematics: it is the \(\large e\)-sequence.
\({\left( {1 + \frac{1}{n}} \right)^n} \uparrow e\;\& \;{\left( {1 + \frac{1}{n}} \right)^{n + 1}} \downarrow e\)
Thus \(C_n=\dfrac{2A_nB_n}{A_n+B_n}\to\dfrac{2e\cdot e}{e+e}=e\).
 
@Aqua1024
Further to my post #8, for the sake of completeness, I'll post the solution I was referring to in post #10.
Please let us know if you came across a nice simple proof.
1623504350351.png
1623504378877.png
 
You must log in or register to reply here.
Top