<s>[INST] Sum $\text{S} = \sum_{i = 2}^{2008}{\frac{1}{a_i}},$ where $a_1 = \frac{1}{3}$ and $ a_{n + 1} = a_n^2 + a_n.$ Define the sequence $\{a_n\}$ where $n \in \mathbb{Z^+}$ given by $a_1 = \frac{1}{3}$ and $$ a_{n + 1} = a_n^2 + a_n.$$ Let $$\text{S} = \sum_{i = 2}^{2008}{\frac{1}{a_i}},$$
then find $\lfloor S \rfloor$ where $\lfloor X \rfloor$ denotes the greatest integer lesser than or equal to $X$.
P.S.: The obvious approach would be to telescope, but as far as I can see, terms do not cancel at all and the estimation of S becomes cumbersome. I have also tried modifying it by adding $\frac{1}{4}$ to both sides and defining $b_n = a_n + \frac{1}{2}$ gives us $$b_{n + 1} = b_n^2 + \frac{1}{4}$$ 
but this doesn't help me in any way to estimate S. One can read off that the original sequence is increasing but i am unable to put an upper bound (such as a G.P.) to find [S].
 [/INST] Start from recurrence relation $a_{n+1} = a_n(a_n+1)$, it is clear if we start from any $a_1 > 0$, $a_n$ will be a strictly increasing sequence.
If for some $N$, we have $a_N = \alpha > 1$, then for all $n \ge N$, we have
$$a_{n+1} = a_{n}(a_{n}+1) \ge a_n(1+\alpha)
\quad\implies\quad \frac{1}{a_{n+1}} \le \frac{1}{a_n(1+\alpha)}$$
This implies for all $k \ge 0$, we have $\displaystyle\;a_{N+k} \le \frac{1}{a_N}\frac{1}{(1+\alpha)^k}$. As a result,
$$\sum_{n=N+1}^\infty \frac{1}{a_n} \le \frac{1}{a_N}\sum_{k=1}^\infty\frac{1}{(1+\alpha)^k}
= \frac{1}{a_N}\frac{\frac{1}{1+\alpha}}{1 - \frac{1}{1+\alpha}} = \frac{1}{a_N\alpha} = \frac{1}{a_N^2}
$$
By brute force, we have
$$(a_1,a_2,a_3,a_4,a_5,\ldots) = (\frac13,\frac49,\frac{52}{81},\frac{6916}{6561},\frac{93206932}{43046721},\ldots)$$
Since $a_n > 1$ start at $n = 4$, we can take $N = 5$. By above argument, we have:
$$5.2182 \sim \sum_{n=2}^5 \frac{1}{a_n}
\le \sum_{n=2}^{2008} \frac{1}{a_n} < \sum_{n=2}^\infty \frac{1}{a_n} \le \sum_{n=2}^5 \frac{1}{a_n} + \frac{1}{a_5^2} \sim 5.4315$$
So the answer is $5$.
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