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If $a_0=1,a_1=1,a_{n+2}=a_{n+1}+a_{n}$, prove the sequence$\left\{\frac{\ln a_n}{\ln a_{n+1}}\right\},\,n\geq 1$ increase.

When $n$ is even this inequality is easy proved by Am-Gm.

If $n$ is even,we have $$\frac{\ln a_n}{\ln a_{n+1}}<\frac{\ln a_{n+1}}{\ln a_{n+2}}\Leftrightarrow \ln a_n\cdot\ln a_{n+2}<(\ln a_{n+1})^2$$

By Am-Gm, and $a_{n-1}a_{n+1}-a_n^2=(-1)^n$, we have$$LHS<\left(\frac{\ln a_n+\ln a_{n+2}}{2}\right)^2=\left(\frac{\ln(a_n\cdot a_{n+2})}{2}\right)^2=\left(\frac{\ln(a_{n+1}^2-1)}{2}\right)^2<RHS.$$

But when n is odd, the inequality is a little difficult, or I didn't use a right method.

Supplement:

Here we prove that $$a_{n-1}a_{n+1}-a_n^2=(-1)^n.$$

It is not hard to prove that if $$A=\begin{pmatrix} 0&1\\ 1&1 \end{pmatrix}$$then$$A^n=\begin{pmatrix} a_{n-1}&a_n\\ a_n&a_{n+1} \end{pmatrix}$$by induction. So $$a_{n-1}a_{n+_1}-a_n^2=\det\begin{vmatrix} 0&1\\ 1&1 \end{vmatrix}^n=\left(\det\begin{vmatrix} 0&1\\ 1&1 \end{vmatrix}\right)^n=(-1)^n$$

Vstal
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1 Answers1

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There is a primary proof by a student in Fudan University(Not me).

First, it is easy to prove that $\left\{\frac{a_{n+1}}{a_{n}}\right\}$ strictly increase.

Second, we can also prove $$0\leq\ln(1+x)\leq x,\quad(\forall\,x\geq 0).$$

And now we begin the proof.

Consider $\forall\,n\geq 3$, we have $$\frac{a_{n+1}}{a_{n-1}}=\frac{a_{n}}{a_{n-1}}\cdot\frac{a_{n+1}}{a_n}\geq\frac{3}{2}\cdot\frac{3}{2}=\frac{9}{4}.$$

So $$ \ln a_{n+1}-\ln a_{n-1}\geq\ln\frac{9}{4}=2\ln\frac{3}{2}>\frac{4}{5}\Rightarrow(\ln a_{n+1}-\ln a_{n-1})^2>\frac{16}{25} $$

By this inequality when $n\geq 5$ we can get $$ \begin{align*} \ln a_{n+1}\cdot\ln a_{n-1}&=\frac{1}{4}(\ln a_{n+1}+\ln a_{n-1})^2-\frac{1}{4} (\ln a_{n+1}-\ln a_{n-1})^2\\ &< \frac{1}{4}(\ln a_{n+1}+\ln a_{n-1})^2-\frac{4}{25}\\ &=\frac{1}{4}(\ln(a_{n+1}\cdot a_{n-1}))^2-\frac{4}{25}\\ &=\frac{1}{4}(\ln(a_{n}^2+(-1)^{n-1}))^2-\frac{4}{25}\\ &=\frac{1}{4}(\ln(a_{n}^2+(-1)^{n-1}))^2-\frac{1}{4}(\ln a_n^2)^2+(\ln a_n)^2-\frac{4}{25}\\ &\leq\frac{1}{4}(\ln(a_n^2+1))^2-\frac{1}{4}(\ln a_n^2)^2+(\ln a_n)^2-\frac{4}{25}\\ &=\frac{1}{4}(\ln(a_n^2+1)-\ln a_n^2)(\ln(a_n^2+1)+\ln a_n^2)+(\ln a_n)^2-\frac{4}{25} \\ &\leq\frac{1}{4}\left(\ln(a_n^2+1)-\ln a_n^2\right)\cdot2\ln((a_n+1)^2)+(\ln a_n)^2-\frac{4}{25}\\ &=\ln(1+\frac{1}{a_n^2})\ln(1+a_n)+(\ln a_n)^2-\frac{4}{25}\\ &\leq \frac{1}{a_n^2}\cdot a_n-\frac{4}{25}+(\ln a_n)^2\\ &\leq\frac{1}{8}-\frac{4}{25}+(\ln a_n)^2\\ &<(\ln a_n)^2 \end{align*} $$

In a word, it is $\ln a_{n-1}\cdot\ln a_{n+1}<(\ln a_n)^2$, which is equal to$$\frac{\ln a_{n-1}}{\ln a_n}<\frac{\ln a_n}{\ln a_{n+1}},\quad n\geq 5.$$

For the rest situation ,we just need to prove it by caculating it.

Calvin Lin
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Vstal
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  • Can you explain more? For example, shouldn't fourth line be $a_{n+1}a_{n-1}=a_n^2\pm1$? – Chrystomath Feb 04 '21 at 09:53
  • @Chrystomath:Sorry, I will add it on the question.Please see the question's supplement. – Vstal Feb 04 '21 at 10:08
  • $\ln a_{n+1}\cdot\ln a_n = \ldots$ should probably be $\ln a_{n+1}\cdot\ln a_{n-1} = \ldots$. – Martin R May 15 '21 at 11:12
  • I've found couple more typos, but overall it seems correct. I took the liberty and fixed all these typos, feel free to check/correct ... (also added few $<$ so that it is clear you got $<$ in the end, not $\leq$) – Sil May 15 '21 at 14:20
  • The first line is false right? $ a_{n+1} / a_n $ is not strictly increasing. It oscillates around the golden ratio. – Calvin Lin May 15 '21 at 15:12
  • -1 Also, $ 1/8 - 4/25 < 0$, so we don't have the penultimate inequality. @Sil Can you review again? – Calvin Lin May 15 '21 at 15:14
  • @CalvinLin That line about strictly increasing $a_{n+1} / a_n$ is false indeed, but I haven't seen it used anywhere. The $1/8 - 4/25 < 0$ seems fine (it gives $(\ln a_n)^2-\frac{7}{200}<(\ln a_n)^2$) – Sil May 15 '21 at 15:22
  • It is used in justifying that $a_{n+1} / a_{n-1} \geq ... \geq 9/4$ . That is a true statement in itself, but needs to be demonstrated in another way. 2) Ah yes, that inequality is fine. Retracting my downvote.
    • – Calvin Lin May 15 '21 at 15:29
  • @CalvinLin Ad 1) Right, that's why I ignored it, i proved it entirely differently :) (it follows quite elementary from definitions...). Agree that OP should mention that part. – Sil May 15 '21 at 15:32