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

#### jacks

##### Well-known member
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

#### castor28

##### Well-known member
MHB Math Scholar
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

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 $\lfloor S\rfloor = 8$.

#### jacks

##### Well-known member
$\displaystyle x_{1} = \frac{1}{3}.$

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

$\displaystyle x_3=\dfrac{52}{81}\in\left(\dfrac{5}{8},\dfrac{2}{3}\right)$.

$\displaystyle x_4>\left(\dfrac{5}{8}\right)^2+\dfrac{5}{8}=\dfrac{65}{64}>1$;

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

$x_5>1^2+1=2$;

$\displaystyle x_5<\left(\dfrac{10}{9}\right)^2+\dfrac{10}{9}=\dfrac{190}{81}<\dfrac{12}{5}$;

$x_6>2^2+2=6$;

$\displaystyle \sum^{2008}_{k=1}\dfrac{1}{x_{k}}>\sum^{5}_{k=1}\dfrac{1}{x_{k}}>+3+\dfrac{9}{4}+\dfrac{3}{2}+\dfrac{9}{10}+\dfrac{5}{12}>8$.

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 $\displaystyle8<\sum^{2008}_{k=1}\dfrac{1}{x_{k}}<9\Longrightarrow\left\lfloor \sum^{2008}_{k=1}\dfrac{1}{x_{k}}\right\rfloor=8$