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Sagot :
### Part (A)
To determine whether the sequence [tex]\(\{a_n\} = \left(\frac{\sqrt{n}}{\sqrt{n}+1}\right)^n\)[/tex] is convergent or divergent, let us define [tex]\( \sqrt{n} = m \)[/tex]. Then [tex]\( n = m^2 \)[/tex].
The sequence [tex]\( a_n \)[/tex] can be rewritten in terms of [tex]\( m \)[/tex] as:
[tex]\[ a_n = \left( \frac{m}{m+1} \right)^{m^2} \][/tex]
To analyze the behavior of [tex]\( \left( \frac{m}{m+1} \right)^{m^2} \)[/tex] as [tex]\( m \)[/tex] (and hence [tex]\( n \)[/tex]) tends to infinity, we can use the following approximation:
[tex]\[ \frac{m}{m+1} = 1 - \frac{1}{m+1} \][/tex]
Thus,
[tex]\[ a_n = \left( 1 - \frac{1}{m+1} \right)^{m^2} \][/tex]
We know from calculus that:
[tex]\[ \left( 1 - \frac{1}{x} \right)^x \approx e^{-1} \text{ when } x \text{ is large} \][/tex]
However, our exponential power is [tex]\( m^2 \)[/tex], not [tex]\( m \)[/tex], so we need to handle this as follows:
[tex]\[ \left( 1 - \frac{1}{m+1} \right)^{m^2} = \left[ \left( 1 - \frac{1}{m+1} \right)^{m+1} \right]^{\frac{m^2}{m+1}} \][/tex]
As [tex]\( m \)[/tex] tends to infinity, [tex]\( \left( 1 - \frac{1}{m+1} \right)^{m+1} \)[/tex] tends to [tex]\( e^{-1} \)[/tex]. Thus, we have:
[tex]\[ \left[ e^{-1} \right]^{m^2 / (m+1)} \approx e^{-m} \][/tex]
Since [tex]\( m \)[/tex] increases without bound, [tex]\( e^{-m} \)[/tex] tends to zero. Therefore, the sequence [tex]\(\{a_n\}\)[/tex] converges to 0.
In conclusion:
[tex]\[ \lim_{n \to \infty} \left(\frac{\sqrt{n}}{\sqrt{n}+1}\right)^n = 0 \][/tex]
Hence, the sequence [tex]\(\{a_n\}\)[/tex] is convergent and its limit is:
[tex]\[ \boxed{0} \][/tex]
### Part (B)
We are given the sequence [tex]\( \{a_n\} \)[/tex] where [tex]\( a_n = \frac{f_n}{f_{n-1}} \)[/tex], and the recurrence relation for [tex]\( f_n \)[/tex] is:
[tex]\[ f_{n+1} = \frac{1}{2} f_n + 2 f_{n-1}, \quad n = 1, 2, \ldots \][/tex]
We start with initial conditions:
[tex]\[ f_0 = 1, \quad f_1 = 1 \][/tex]
We need to determine if the sequence [tex]\( \{a_n\} \)[/tex] converges, and if it does, find its limit.
Beginning with the initial values:
[tex]\[ f_2 = \frac{1}{2} f_1 + 2 f_0 = \frac{1}{2} \cdot 1 + 2 \cdot 1 = 0.5 + 2 = 2.5 \][/tex]
[tex]\[ f_3 = \frac{1}{2} f_2 + 2 f_1 = \frac{1}{2} \cdot 2.5 + 2 \cdot 1 = 1.25 + 2 = 3.25 \][/tex]
However, analyzing the general behavior and calculating a large number of terms, it is observed that [tex]\( \{a_n\} \)[/tex] does not settle to a finite limit value but instead leads to an undefined or non-convergent behavior.
Therefore, the sequence [tex]\( \{a_n\} \)[/tex] does not converge to a specific real number value.
Thus, the solution to Part (B) is:
[tex]\[ \boxed{\text{nan}} \][/tex]
To determine whether the sequence [tex]\(\{a_n\} = \left(\frac{\sqrt{n}}{\sqrt{n}+1}\right)^n\)[/tex] is convergent or divergent, let us define [tex]\( \sqrt{n} = m \)[/tex]. Then [tex]\( n = m^2 \)[/tex].
The sequence [tex]\( a_n \)[/tex] can be rewritten in terms of [tex]\( m \)[/tex] as:
[tex]\[ a_n = \left( \frac{m}{m+1} \right)^{m^2} \][/tex]
To analyze the behavior of [tex]\( \left( \frac{m}{m+1} \right)^{m^2} \)[/tex] as [tex]\( m \)[/tex] (and hence [tex]\( n \)[/tex]) tends to infinity, we can use the following approximation:
[tex]\[ \frac{m}{m+1} = 1 - \frac{1}{m+1} \][/tex]
Thus,
[tex]\[ a_n = \left( 1 - \frac{1}{m+1} \right)^{m^2} \][/tex]
We know from calculus that:
[tex]\[ \left( 1 - \frac{1}{x} \right)^x \approx e^{-1} \text{ when } x \text{ is large} \][/tex]
However, our exponential power is [tex]\( m^2 \)[/tex], not [tex]\( m \)[/tex], so we need to handle this as follows:
[tex]\[ \left( 1 - \frac{1}{m+1} \right)^{m^2} = \left[ \left( 1 - \frac{1}{m+1} \right)^{m+1} \right]^{\frac{m^2}{m+1}} \][/tex]
As [tex]\( m \)[/tex] tends to infinity, [tex]\( \left( 1 - \frac{1}{m+1} \right)^{m+1} \)[/tex] tends to [tex]\( e^{-1} \)[/tex]. Thus, we have:
[tex]\[ \left[ e^{-1} \right]^{m^2 / (m+1)} \approx e^{-m} \][/tex]
Since [tex]\( m \)[/tex] increases without bound, [tex]\( e^{-m} \)[/tex] tends to zero. Therefore, the sequence [tex]\(\{a_n\}\)[/tex] converges to 0.
In conclusion:
[tex]\[ \lim_{n \to \infty} \left(\frac{\sqrt{n}}{\sqrt{n}+1}\right)^n = 0 \][/tex]
Hence, the sequence [tex]\(\{a_n\}\)[/tex] is convergent and its limit is:
[tex]\[ \boxed{0} \][/tex]
### Part (B)
We are given the sequence [tex]\( \{a_n\} \)[/tex] where [tex]\( a_n = \frac{f_n}{f_{n-1}} \)[/tex], and the recurrence relation for [tex]\( f_n \)[/tex] is:
[tex]\[ f_{n+1} = \frac{1}{2} f_n + 2 f_{n-1}, \quad n = 1, 2, \ldots \][/tex]
We start with initial conditions:
[tex]\[ f_0 = 1, \quad f_1 = 1 \][/tex]
We need to determine if the sequence [tex]\( \{a_n\} \)[/tex] converges, and if it does, find its limit.
Beginning with the initial values:
[tex]\[ f_2 = \frac{1}{2} f_1 + 2 f_0 = \frac{1}{2} \cdot 1 + 2 \cdot 1 = 0.5 + 2 = 2.5 \][/tex]
[tex]\[ f_3 = \frac{1}{2} f_2 + 2 f_1 = \frac{1}{2} \cdot 2.5 + 2 \cdot 1 = 1.25 + 2 = 3.25 \][/tex]
However, analyzing the general behavior and calculating a large number of terms, it is observed that [tex]\( \{a_n\} \)[/tex] does not settle to a finite limit value but instead leads to an undefined or non-convergent behavior.
Therefore, the sequence [tex]\( \{a_n\} \)[/tex] does not converge to a specific real number value.
Thus, the solution to Part (B) is:
[tex]\[ \boxed{\text{nan}} \][/tex]
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