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

jacks

Well-known member
Apr 5, 2012
226
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
Oct 18, 2017
245
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
Apr 5, 2012
226
$\displaystyle x_{1} = \frac{1}{3}.$

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

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


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

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


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

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


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


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


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