Monotony of sequence

Dec 2015
1,082
169
Earth
Is (*) correct ?
If \(\displaystyle a_n \uparrow \) then \(\displaystyle \dfrac{1}{a_n } \downarrow \; \) ; \(\displaystyle m_n =(a_{n+1}-a_n )(a_{n+1}^{-1} -a_{n}^{-1} ) \leq 0\).

\(\displaystyle m_n = 1-a_{n+1}a_{n}^{-1} -a_n a_{n+1}^{-1} +1 =2-a_{n+1}a_{n}^{-1} -a_n a_{n+1}^{-1}\leq 0\).

\(\displaystyle m_{n+1}-m_{n}=2-a_{n+2}a_{n+1}^{-1} -a_{n+1} a_{n+2}^{-1}-[2-a_{n+1}a_{n}^{-1} -a_n a_{n+1}^{-1}]=a_{n+1}a_{n}^{-1} +a_n a_{n+1}^{-1} - a_{n+2}a_{n+1}^{-1} -a_{n+1} a_{n+2}^{-1}\leq 0\).

\(\displaystyle a_{n+1}a_{n}^{-1} +a_n a_{n+1}^{-1} \leq a_{n+2}a_{n+1}^{-1} + a_{n+1} a_{n+2}^{-1} \).

\(\displaystyle \dfrac{a_{n+1}}{a_n }+\dfrac{a_n }{a_{n+1}} \leq \dfrac{a_{n+2}}{a_{n+1} }+\dfrac{a_{n+1} }{a_{n+2}}\leq \dfrac{a_{n+3}}{a_{n+2} }+\dfrac{a_{n+2} }{a_{n+3}} \leq \dotsc \leq \dfrac{a_{2n}}{a_{2n-1} }+\dfrac{a_{2n-1} }{a_{2n}} \).

(*) \(\displaystyle \dfrac{a_{n+1}}{a_n }+\dfrac{a_n }{a_{n+1}} \leq \dfrac{a_{2n}}{a_{2n-1} }+\dfrac{a_{2n} }{a_{2n-1}}\).

Example: \(\displaystyle a_n = n!\) then by (*) : \(\displaystyle n+1+\dfrac{1}{n+1} \leq 2n +\dfrac{1}{2n}\).
 
Last edited:
Dec 2015
1,082
169
Earth
(*) \(\displaystyle \displaystyle \dfrac{a_{n+1}}{a_n }+\dfrac{a_n }{a_{n+1}} \leq \dfrac{a_{2n}}{a_{2n-1} }+\dfrac{a_{2n-1} }{a_{2n}} \).

Example2 : \(\displaystyle a_n = \sqrt[n]{n} \) then by (*) : \(\displaystyle \dfrac{\sqrt[n+1]{n+1}}{\sqrt[n]{n}}+\dfrac{\sqrt[n]{n}}{\sqrt[n+1]{n+1}} \leq \dfrac{\sqrt[2n]{2n}}{\sqrt[2n-1]{2n-1}} + \dfrac{\sqrt[2n-1]{2n-1}}{\sqrt[2n]{2n}}\).
 
Dec 2015
1,082
169
Earth
It's all elementary simple math .
Let R define the relation over 0.
\(\displaystyle ( \cdot ) \begin{cases} a_{n+1}-a_n \; R \; 0 \\ \\ \dfrac{1}{a_{n+1}}-\dfrac{1}{a_n } \; R^{-} \; 0 \end{cases} \; \Rightarrow [a_{n+1} - a_n ][\dfrac{1}{a_{n+1}} -\dfrac{1}{a_n } ] \; \leq 0 \).

If \(\displaystyle a_n \uparrow \) then \(\displaystyle a_{n}^{-1} \downarrow\) and if \(\displaystyle a_n \downarrow \) then \(\displaystyle a_{n}^{-1} \uparrow \).

\(\displaystyle RR^{-} \equiv \: \leq 0 \).
 
Last edited:
Oct 2018
129
96
USA
(*) Doesn't seem to be true for $a_n = n$, unless i'm missing something.

(*) $\frac{n+1}{n} + \frac{n}{n+1} \leq \frac{2n}{2n-1} + \frac{2n-1}{2n}$

$\implies \frac{(n+1)^2 + n^2}{n(n+1)} \leq \frac{4n^2+(2n-1)^2}{2n(2n-1)}$

$\implies 2+\frac{1}{n^2+n} \leq 2+\frac{1}{4n^2-2n}$

$\implies 4n^2 - 2n \leq n^2 + n$

$\implies n^2 - n \leq 0$

$\implies n^2 \leq n$

Which is only true for $n=1$
 
  • Like
Reactions: idontknow
Dec 2015
1,082
169
Earth
Check post#3 and see if you can arrive to something similiar to post#4 .
\(\displaystyle
[a_{n+1} - a_n ][\dfrac{1}{a_{n+1}} -\dfrac{1}{a_n } ] =2-\dfrac{a(n)}{a(n+1)}-\dfrac{a(n+1)}{a(n)} \; \leq 0

\).

example1 : \(\displaystyle a(n)=n \) , \(\displaystyle \; 2\leq \dfrac{n}{n+1}+\dfrac{n+1}{n}\).

example2 : \(\displaystyle a(n)=n! \) , \(\displaystyle \; 2 \leq \dfrac{1}{n+1}+1+n\).

example3 : \(\displaystyle a(n)=\sqrt{n} \) , \(\displaystyle \; 2 \leq \sqrt{\dfrac{n}{n+1}}+\sqrt{\dfrac{n+1}{n}}\).

example4 : \(\displaystyle a(n)=n^n \) , \(\displaystyle \; 2 \leq \dfrac{n^n }{(n+1)^{n+1}}+\dfrac{(n+1)^{n+1}}{n^n }\).
 
Last edited:
Dec 2015
1,082
169
Earth
Post#5 just gives the AM-GM : \(\displaystyle \dfrac{\frac{a(n)}{a(n+1)}+\frac{a(n+1)}{a(n)}}{2} \geq \sqrt[4]{1}=1\).

\(\displaystyle \frac{a(n)}{a(n+1)}+\frac{a(n+1)}{a(n)} \geq 2 \).

But \(\displaystyle m_n = 1-a_{n+1}a_{n}^{-1} -a_n a_{n+1}^{-1} +1 =2-a_{n+1}a_{n}^{-1} -a_n a_{n+1}^{-1}\leq 0 \) and \(\displaystyle |m_{n+p}-m_n | \) may give inequalities different of AM-GM.