# Monotony of sequence

#### idontknow

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 (*) : $$\displaystyle n+1+\dfrac{1}{n+1} \leq 2n +\dfrac{1}{2n}$$.

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#### idontknow

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

Example2 : $$\displaystyle a_n = \sqrt[n]{n}$$ then by (*) : $$\displaystyle \dfrac{\sqrt[n+1]{n+1}}{\sqrt[n]{n}}+\dfrac{\sqrt[n]{n}}{\sqrt[n+1]{n+1}} \leq \dfrac{\sqrt[2n]{2n}}{\sqrt[2n-1]{2n-1}} + \dfrac{\sqrt[2n-1]{2n-1}}{\sqrt[2n]{2n}}$$.

#### idontknow

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 $$\displaystyle a_{n}^{-1} \uparrow$$.

$$\displaystyle RR^{-} \equiv \: \leq 0$$.

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#### Greens

(*) 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 $n=1$

idontknow

#### idontknow

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

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#### idontknow

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.