orl wrote: Let $ \{a_n\}^{\infty}_1$ be a sequence of real numbers such that $ a_1 = 2,$ and \[ a_{n + 1} = a^2_n - a_n + 1, \forall n \in \mathbb{N}. \] Prove that \[ 1 - \frac {1}{2003^{2003}} < \sum^{2003}_{i = 1} \frac {1}{a_i} < 1. \] $ a_{n + 1} = a^2_n - a_n + 1$ thus $ \frac{1}{a_n}=\frac{1}{a_n-1}-\frac{1}{a_{n+1}-1}$ hence $ \sum^{2003}_{i = 1} \frac {1}{a_i}=\frac{1}{a_1-1}-\frac{1}{a_{2004}-1}$ The last is easy

How is the left side of the inequality true?

Why do you say it trivial sir???? I havenot seen any thing that yoy say trivial?

arkanm wrote: LHS inequality is saying that $1-\frac{1}{2003^{2003}}<\frac{1}{a_1-1}-\frac{1}{a_{2004}-1}\Leftrightarrow a_{2004}-1>2003^{2003}$. Now from the recursion observe that $a_{n+1}-1=a_n(a_n-1)$, hence by repeatedly applying the recursion to $a_n-1$ we get $a_{n+1}-1=(a_1-1)\prod_{i=2}^na_i=\prod_{i=2}^na_i$. It suffices to show that $\prod_{i=2}^{2003}a_i>2003^{2003}$, which is trivial. First, your indexing is wrong; it should be $a_{n+1} - 1 = \prod_{i=1}^na_i$; but that is mostly irrelevant. What is relevant is your claim that $\prod_{i=1}^{2003} a_i> 2003^{2003}$ is trivial, which it isn't (at least at first sight). We want to show that $a_{n+1} > n^n + 1$. This can easily be checked for the first few small values of $n$. The function $f(x) = x^2 - x + 1$ is increasing on $[1,\infty)$, and we have $a_{n+2} = f(a_{n+1})$, so $a_{n+2} = f(a_{n+1}) > f(n^n+1) = n^{2n} +n^n +1$, which we would like to be at least $(n+1)^{n+1} +1$. But $n^{2n} > (n+1)^{n+1}$ is equivalent to $\dfrac {n^n}{n+1} > \left (\dfrac {n+1} {n}\right )^n = \left (1 + \dfrac {1} {n}\right )^n$. Now, $LHS > 3$ for $n\geq 3$, and $RHS < 3$ (from the known proof for $\textrm{e}$), so induction does the job.

See also here.

Funny problem The LHS bound is pretty stupid and somewhat of a red herring. Notice that $$a_{n+1} = a_n(a_n-1) \iff \frac 1{a_n} = \frac 1{a_n-1} - \frac 1{a_{n+1}-1}.$$As a result $$\frac 1{a_1}+\frac 1{a_2}+\cdots + \frac 1{a_{2003}} = 1 - \frac 1{a_{2004} - 1}.$$The first display yields further that $a_{n+1}-1 > (a_n-1)^2$, so recursing $$a_{2004} - 1 > (a_2-1)^{2^{2002}} = 2^{2^{2002}} > (2^{2003})^{2003} > 2003^{2003},$$and obviously $a_{2004} - 1 > 0$, proving both bounds.