MHB Evaluate ⌊ 1/a_1+1/a_2+....+1/a_{2008} ⌋

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The sequence \( a_n \) is defined recursively with \( a_1 = \frac{1}{3} \) and \( a_{k+1} = a_k^2 + a_k \). The sum \( S = \sum_{k=1}^{2008} \frac{1}{a_k} \) is evaluated, with initial terms calculated to show \( S(5) = 8.2182 \). An upper bound for the remaining terms \( T(6) < 0.1708 \) leads to the conclusion that \( S < 8.3890 \). Consequently, it is established that \( \lfloor S \rfloor = 8 \). The rapid growth of the sequence ensures that the sum remains bounded between 8 and 9.
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Consider the sequence $a_{n}$ given by $\displaystyle a_{1} = \frac{1}{3}$ and $\displaystyle a_{k+1}=a^2_{k}+a_{k}$ for $k\geq 2$and Let $\displaystyle S = \frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots \cdots +\frac{1}{a_{2008}}$. Then $\lfloor S \rfloor $ is
 
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jacks said:
Consider the sequence $a_{n}$ given by $\displaystyle a_{1} = \frac{1}{3}$ and $\displaystyle a_{k+1}=a^2_{k}+a_{k}$ for $k\geq 2$and Let $\displaystyle S = \frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots \cdots +\frac{1}{a_{2008}}$. Then $\lfloor S \rfloor $ is
[sp]
Note: I assume that the recurrence holds for $k\ge\bf1$.

Let us write $ b_n = \dfrac{1}{a_n}$, $S(n) = \sum_{k=1}^nb_k$, and $T(n) = \sum_{k=n}^{2008}b_k$. As $a_{n+1} > a_n^2$, we have $b_{n+1} < b_n^2$. If $b_n < 1$, comparison with the geometric progression of ratio $b_n$ gives:
$$T(n) < \frac{b_n}{1-b_n}.$$

We compute now the first few terms of the series:
$$
\begin{array}{c|c|c|c}
n&a_n&b_n&S(n)\\
\hline
1&0.3333&3.0000&3.0000\\
2&0.4444&2.2500&5.2500\\
3&0.6420&1.5577&6.8077\\
4&1.0541&0.9487&7.7564\\
5&2.1653&0.4618&8.2182\\
6&6.8536&0.1459&8.3641\\
\end{array}
$$
Using the remark above, we find $T(6) < \dfrac{b_6}{1-b_6} = 0.1708$. This shows that:
$$
S(5) = 8.2182 < S = S(2008) = S(5) + T(6) < 8.2182 + 0.1708 = 8.3890
$$
and therefore $\lfloor S\rfloor = 8$.
[/sp]
 
[sp]$\displaystyle x_{1} = \frac{1}{3}.$

$\displaystyle x_2=\dfrac{4}{9}$.

$\displaystyle x_3=\dfrac{52}{81}\in\left(\dfrac{5}{8},\dfrac{2}{3}\right)$.$\displaystyle x_4>\left(\dfrac{5}{8}\right)^2+\dfrac{5}{8}=\dfrac{65}{64}>1$;

$\displaystyle x_4<\left(\dfrac{2}{3}\right)^2+\dfrac{2}{3}=\dfrac{10}{9}$;$x_5>1^2+1=2$;

$\displaystyle x_5<\left(\dfrac{10}{9}\right)^2+\dfrac{10}{9}=\dfrac{190}{81}<\dfrac{12}{5}$;$x_6>2^2+2=6$;$\displaystyle \sum^{2008}_{k=1}\dfrac{1}{x_{k}}>\sum^{5}_{k=1}\dfrac{1}{x_{k}}>+3+\dfrac{9}{4}+\dfrac{3}{2}+\dfrac{9}{10}+\dfrac{5}{12}>8$.Beginning from $x_6$, the sequence $x_n$ is growing very fast.

For $k\ge 6: x_k>6^{k-5}$.

$\displaystyle \sum_{k=6}^{2008}\dfrac{1}{x_k}<\sum_{m=1}^{+\infty}\dfrac{1}{6^m}=\dfrac{1}{5}$.Results: $\displaystyle \sum^{2008}_{k=2}\dfrac{1}{x_{k}}=\sum^{5}_{k=1}\dfrac{1}{x_{k}}+\sum^{2008}_{k=6}\dfrac{1}{x_{k}}<+3+\dfrac{9}{4}+\dfrac{8}{5}+1+\dfrac{1}{2}+\dfrac{1}{5}<9$.So we have $\displaystyle8<\sum^{2008}_{k=1}\dfrac{1}{x_{k}}<9\Longrightarrow\left\lfloor \sum^{2008}_{k=1}\dfrac{1}{x_{k}}\right\rfloor=8$[/sp]
 
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