Let $\{a_n\}$ be a sequence of real numbers such that $a_1=2$, $a_{n+1} = a_n^2 -a_n+1$, for $n=1,2,3..$. [Cont] 
Let $\{a_n\}$ be a sequence of real numbers such that $a_1=2$, $a_{n+1} = a_n^2 -a_n+1$, for $n=1,2,3..$. Let $S=\frac{1}{a_1}+\frac{1}{a_2} ....+\frac{1}{a_{2018}}$, then prove that



*

*$S>1-\frac{1}{2018^{2018}}$




*$S<1$




*$S>1-\frac{1}{2017^{2017}}$

$$a_{n+1}-1 = a_n^2-1-a_n+1$$
$$a_{n+1} -1 = (a_n-1)(a_n)$$
$$\frac{1}{a_{n+1}-1} =\frac{1}{a_n-1} -\frac{1}{a_n}$$
$$\frac{1}{a_n}=\frac{1}{a_n-1} -\frac{1}{a_{n+1}-1}$$
So $$S=1-\frac{1}{a_{2019}-1}$$
How do I proceed from here?

You have already shown that
$S = 1-\frac{1}{a_{2019}-1}$
So, $S \gt 1-\frac{1}{2018^{2018}}$ if $\, a_{2019} \gt {2018}^{2018} + 1$
By induction -
As $a_{2} = {a_1}^2 - a_1 + 1 = 3 \gt 1^1 + 1$
and $a_{3} = {a_2}^2 - a_2 + 1 = 7 \gt 2^2 + 1$
For a value of $n \ge 3$,
If $a_{n+1} \gt n^n + 1$, we need to show that $a_{n+2} \gt (n+1)^{n+1} + 1$
$a_{n+2} \ge (n^n + 1)^2 - (n^n+1) + 1 = n^{2n} + n^n + 1 \gt (n+1)^{n+1} + 1$
which is true if we can show that $n^{2n} \gt (n+1)^{n+1}$
or if we can show that $\frac {n^n} {n+1} \gt (\frac{n+1}{n})^n$ for $n \ge 3$
For which please see the below links -
The first answer in the first link beautifully shows that
$(\frac{n+1}{n})^n \lt n$ for $n \ge 3$
Prove by induction that for all $n \geq 3$: $n^{n+1} > (n+1)^n$
Prove $(\frac{n+1}{n})^{n+1}$ is decreasing
This shows for prove $(\frac{n+1}{n})^n < (\frac{n}{n-1})^n$ if $n\geq2$
