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

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  • Thread starter juantheron
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In summary, the sequence $a_{n}$ is given by $\displaystyle a_{1} = \frac{1}{3}$ and $\displaystyle a_{k+1}=a^2_{k}+a_{k}$ for $k\geq 2$, and the sum of the reciprocals of the first 2008 terms is less than 9 and greater than 8, resulting in $\lfloor S\rfloor = 8$.
  • #1
juantheron
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1
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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  • #2
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]
 
  • #3
[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]
 

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

1. How do you evaluate the expression ⌊ 1/a_1+1/a_2+....+1/a_{2008} ⌋?

To evaluate this expression, you need to first find the values of a_1, a_2, ..., a_{2008}. Then, plug these values into the expression and perform the necessary arithmetic operations. The resulting value will be the evaluated expression.

2. What does the symbol ⌊ ⌋ mean in this expression?

The symbol ⌊ ⌋ represents the floor function, which rounds a given number down to the nearest integer. In this expression, it is used to indicate that the result should be an integer.

3. Can the expression ⌊ 1/a_1+1/a_2+....+1/a_{2008} ⌋ be simplified?

It depends on the values of a_1, a_2, ..., a_{2008}. In some cases, the expression can be simplified by finding a common denominator and combining the fractions. However, in other cases, the expression may not be able to be simplified any further.

4. What is the significance of evaluating this expression?

This expression is commonly used in mathematics and statistics to calculate the harmonic mean, which is a type of average. It is also used in various mathematical proofs and calculations.

5. Are there any special cases or restrictions for the values of a_1, a_2, ..., a_{2008} in this expression?

Yes, the values of a_1, a_2, ..., a_{2008} cannot be equal to zero, as this would result in an undefined expression. Additionally, if any of the values are negative, the expression may not be able to be evaluated.

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