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Please do check if my solution is correct. Consider the sequence $\{a_n\}$ defined recursively by $a_1 = \frac{3}{2}$ and $a_{n+1} = a_n^2 -2a_n +2, \quad n\ge 1.$

Show that the sequence is decreasing. Here is my attempt.

It suffices to show that $a_n- a_{n+1} > 0, n\ge 1$.

$$a_n- a_{n+1}=a_n-[a_n^2 -2a_n +2]= -a_n^2+3a_n-2$$ $$=-[a_n^2-3a_n+2]=-[a_n-2][a_n-1]$$

Observe that $a_n-2 <0$ for $n\ge 1$ and $a_n-1 > 0$ for $n>1$, then the product $[a_n-2][a_n-1] <0$ which implies that $-[a_n-2][a_n-1]$ is positive.

Hence $$a_n- a_{n+1}=-[a_n^2-3a_n+2]=-[a_n-2][a_n-1]>0.$$ Do I make any sense? Thanks for helping me.

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    `Observe that a_n − 2 < 0 for n ≥ 1 and a_n − 1 > 0 for n > 1`. Both are true, but you can't just state them, you need to prove them. Hint: the initial condition $a_1 = 3/2$ does matter. – 2017-01-27

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It appears $a_{n+1}=(a_n-1)^2+1$ so that $a_{n+k} = (a_n-1)^{2^k}+1$, showing that $a_n= 1+2^{-2^{n-1}}$ which is clearly decreasing.

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    :-) Was just going to ask you if you checked that. – 2017-01-27
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    @Jaynot don't forget to participate in the community by upvoting answers you find helpful. It helps future people know what answers are found useful when searching. – 2017-01-27
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    How do I do that please @Adam Hughes because I cant find any option here that says I can do that? – 2017-01-27
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    Hey @Jaynot just click the little arrows pointing up next to the questions. – 2017-01-27
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    Lol. @Jaynot Also, if you find an answer that fully answers your question, hit the little checkmark button below the arrows. – 2017-01-27
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    @Jaynot my pleasure, I figured you probably just didn't know. :) – 2017-01-27
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    Lol. Says thank you with no vote-activity? Yeah, probably didn't know. Think we should point out the tour page and stuff @AdamHughes ? – 2017-01-27
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    Lol. A cruelly deserved policy you run. – 2017-01-27
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I am not so sure why you wouldn't do this, but prove by induction that with $a_1=\frac32$, then

$$a_n=1+\frac1{2^{2^{n-1}}}$$

whereupon showing that it is monotone decreasing is easy.