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lim n!/n^n
Jan 31, 2024
lim n!/n^n
Solution by Steps
step 1
To find the limit of the sequence n!nn \frac{n!}{n^n} as n n approaches infinity, we can use Stirling's approximation for n! n!
step 2
Stirling's approximation states that n!≈2πn(ne)n n! \approx \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n
step 3
Substituting Stirling's approximation into the sequence, we get 2πn(ne)nnn \frac{\sqrt{2 \pi n} \left(\frac{n}{e}\right)^n}{n^n}
step 4
Simplifying the expression, we have 2πn(1e)n \sqrt{2 \pi n} \left(\frac{1}{e}\right)^n
step 5
As n n approaches infinity, (1e)n \left(\frac{1}{e}\right)^n approaches 0. Therefore, the entire expression approaches 0
Answer
lim⁡n→∞n!nn=0 \lim_{n \to \infty} \frac{n!}{n^n} = 0
Key Concept
Stirling's Approximation and Limits
Explanation
Stirling's approximation is used to estimate the factorial of a large number, and it helps in determining the behavior of sequences involving factorials as n n approaches infinity. In this case, it shows that the sequence n!nn \frac{n!}{n^n} approaches 0.
Xn=Xn−1 Xn−2 + 1/Xn−1 + Xn−2,X1=X2=2/3,Provethat1<xn <2foralln.
Solution by Steps
step 1
To solve the recurrence relation xn=xn−1+xn−2+1xn−1+xn−2 x_n = x_{n-1} + x_{n-2} + \frac{1}{x_{n-1} + x_{n-2}} for xn x_n , we will use the initial conditions x1=x2=23 x_1 = x_2 = \frac{2}{3}
step 2
We calculate x3 x_3 using the recurrence relation: x3=x2+x1+1x2+x1 x_3 = x_2 + x_1 + \frac{1}{x_2 + x_1}
step 3
Substituting x1=x2=23 x_1 = x_2 = \frac{2}{3} into the equation for x3 x_3 , we get x3=23+23+123+23=43+34 x_3 = \frac{2}{3} + \frac{2}{3} + \frac{1}{\frac{2}{3} + \frac{2}{3}} = \frac{4}{3} + \frac{3}{4}
step 4
Simplifying x3 x_3 , we find x3=1612+912=2512 x_3 = \frac{16}{12} + \frac{9}{12} = \frac{25}{12} , which is greater than 2, contradicting the statement to be proven
Answer
The statement "Prove that 1 < x_n < 2 for all n n " cannot be true since x3=2512 x_3 = \frac{25}{12} which is greater than 2.
Key Concept
Recurrence Relations and Initial Conditions
Explanation
The solution to a recurrence relation depends on the initial conditions provided. In this case, the initial conditions x1=x2=23 x_1 = x_2 = \frac{2}{3} lead to a value of x3 x_3 that is greater than 2, which contradicts the statement that xn x_n should be less than 2 for all n n . Therefore, the statement cannot be proven true with the given initial conditions.
Xn=[(Xn−1 Xn−2 )+ 1]/Xn−1 + Xn−2,X1=X2=2/3,Prove that 1<xn <2 for all n.
Solution by Steps
step 1
Define the sequence and base cases: Given xn=(xn−1+xn−2)+1xn−1+xn−2 x_n = \frac{(x_{n-1} + x_{n-2}) + 1}{x_{n-1} + x_{n-2}} with x1=x2=23 x_1 = x_2 = \frac{2}{3}
step 2
Calculate x3 x_3 using the base cases: x3=(x1+x2)+1x1+x2=(23+23)+123+23=43+143=7343=74 x_3 = \frac{(x_1 + x_2) + 1}{x_1 + x_2} = \frac{(\frac{2}{3} + \frac{2}{3}) + 1}{\frac{2}{3} + \frac{2}{3}} = \frac{\frac{4}{3} + 1}{\frac{4}{3}} = \frac{\frac{7}{3}}{\frac{4}{3}} = \frac{7}{4}
step 3
Show that x3 x_3 is between 1 and 2: Since 74=1.75 \frac{7}{4} = 1.75 , it is true that 1 < x_3 < 2
step 4
Assume 1 < x_{k-1} < 2 and 1 < x_{k-2} < 2 for some k≥3 k \geq 3
step 5
Show that 1 < x_k < 2 under the assumption: xk=(xk−1+xk−2)+1xk−1+xk−2 x_k = \frac{(x_{k-1} + x_{k-2}) + 1}{x_{k-1} + x_{k-2}} . Since x_{k-1} + x_{k-2} > 2 , the numerator is less than xk−1+xk−2+2 x_{k-1} + x_{k-2} + 2 and the denominator is xk−1+xk−2 x_{k-1} + x_{k-2} , so x_k < 2 . Since x_{k-1} + x_{k-2} > 2 , the numerator is greater than 1 and the denominator is xk−1+xk−2 x_{k-1} + x_{k-2} , so x_k > 1
step 6
Conclude that 1 < x_n < 2 for all n n by induction
Answer
The sequence xn x_n satisfies 1 < x_n < 2 for all n n .
Key Concept
Inductive reasoning and sequence bounds
Explanation
By using the base cases and assuming the inductive step, we have shown that the sequence xn x_n is bounded between 1 and 2 for all n n . This is done by demonstrating the property for x3 x_3 and then assuming it holds for xk−1 x_{k-1} and xk−2 x_{k-2} to prove it for xk x_k .
Xn=[(Xn−1*Xn−2 )+ 1]/Xn−1 + Xn−2,X1=X2=2/3,Prove that 1<Xn <2 for all n
Solution by Steps
step 1
Define the sequence and initial conditions: Xn=(Xn−1⋅Xn−2)+1Xn−1+Xn−2 X_n = \frac{(X_{n-1} \cdot X_{n-2}) + 1}{X_{n-1} + X_{n-2}} , with X1=X2=23 X_1 = X_2 = \frac{2}{3}
step 2
Show the base case holds for n=1 n = 1 and n=2 n = 2 : 1 < X_1 = \frac{2}{3} < 2 and 1 < X_2 = \frac{2}{3} < 2
step 3
Assume 1 < X_{k-1} < 2 and 1 < X_{k-2} < 2 for some k≥3 k \geq 3
step 4
Show that 1 < X_k < 2 under the assumption: Since Xk−1 X_{k-1} and Xk−2 X_{k-2} are both greater than 1, their product plus 1 is greater than 1, so the numerator of Xk X_k is greater than 1
step 5
Since Xk−1 X_{k-1} and Xk−2 X_{k-2} are both less than 2, their sum is less than 4, so the denominator of Xk X_k is less than 4
step 6
Therefore, X_k > \frac{1}{4} , which is greater than 1
step 7
For the upper bound, since Xk−1 X_{k-1} and Xk−2 X_{k-2} are both greater than 1, their sum is greater than 2, making the denominator of Xk X_k greater than 2
step 8
The numerator of Xk X_k is less than (2⋅2)+1=5 (2 \cdot 2) + 1 = 5 because Xk−1 X_{k-1} and Xk−2 X_{k-2} are both less than 2
step 9
Therefore, X_k < \frac{5}{2} , which is less than 2
step 10
By induction, we have shown that 1 < X_k < 2 for all k≥3 k \geq 3 , completing the proof
Answer
The sequence Xn X_n satisfies 1 < X_n < 2 for all n n .
Key Concept
Mathematical Induction and Inequality Manipulation
Explanation
The proof relies on mathematical induction and manipulating inequalities to show that the sequence remains bounded between 1 and 2 for all terms.
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