Investigating the Convergence of the Sequence aₙ = (2n – 1)! / (2n + 1)!
We are investigating the convergence of a sequence defined by a specific recurrence relation. We’ll break down the process into clear steps.
1. Simplifying the Expression
Starting with the given expression, we want to simplify it to a more manageable form:
aₙ = (2n – 1)! / (2n + 1)!
We can rewrite (2n + 1)! by recognizing the pattern of the factorial function:
(2n + 1)! = (2n + 1) * (2n + 1 – 1) * (2n + 1 – 1 – 1) * … * 3 * 2 * 1
Notice that we are subtracting 1 each time as we write the product. This pattern continues until we reach 1.
Now, we can see that part of this expression is exactly (2n – 1)!:
(2n – 1)! = (2n – 1) * (2n – 2) * … * 3 * 2 * 1
So we can substitute this back into our original expression:
aₙ = (2n – 1)! / ((2n + 1) * (2n) * (2n – 1)!)
Since the (2n – 1)! terms cancel out, we are left with a simpler expression:
aₙ = 1 / ((2n + 1) * (2n))
This step helps us see the structure of the sequence more clearly by recognizing the factorial pattern and canceling out common terms.
2. Finding the Limit
Now, we want to find the limit of the sequence as n approaches infinity. This will tell us if the sequence converges:
lim (n → ∞) aₙ = lim (n → ∞) 1 / ((2n + 1) * (2n))
As n becomes very large, the denominator grows much faster than the numerator, causing the fraction to approach 0:
lim (n → ∞) 1 / (4n² + 2n) = 0
This step confirms that the sequence gets arbitrarily close to 0 as n gets larger.
3. Conclusion
Since the limit of the sequence is 0, the sequence converges, and the limit is 0. This investigation has taken us through the process of simplifying a complex expression, understanding its behavior, and determining its convergence. It illustrates the power of mathematical analysis in understanding sequences and their properties.
Investigating the Convergence of the Sequence \( aₙ = n^{1/n} \)
We are investigating the convergence of a sequence defined by the expression \( aₙ = n^{1/n} \). We’ll break down the process into clear steps.
1. Understanding the Expression
The expression \( aₙ = n^{1/n} \) represents the nth root of n. As n increases, we are taking higher and higher roots of larger and larger numbers. We want to understand how this behaves as n approaches infinity.
2. Relating the Expression to the Exponential Function
We can rewrite any expression of the form \( x^y \) as \( e^{y \ln x} \). Applying this to our expression, we have:
\( aₙ = n^{1/n} = e^{(1/n) \ln n} \)
This step helps us relate the original expression to the exponential function, which has well-known properties and behavior.
3. Finding the Limit
To find the limit of the sequence as n approaches infinity, we can now work with the equivalent expression:
\( \lim_{n \to ∞} e^{(1/n) \ln n} \)
Since the exponential function is continuous, we can move the limit inside:
\( \lim_{n \to ∞} e^{(1/n) \ln n} = e^{\lim_{n \to ∞} (1/n) \ln n} \)
By applying L’Hôpital’s rule to the expression inside the limit, we find:
\( \lim_{n \to ∞} (1/n) \ln n = 0 \)
So the original limit becomes:
\( \lim_{n \to ∞} n^{1/n} = e^0 = 1 \)
4. Conclusion
The sequence \( aₙ = n^{1/n} \) converges to 1 as n approaches infinity. This result is a fascinating insight into the behavior of taking roots of increasing numbers. It shows that as we take higher and higher roots of larger and larger numbers, the values approach 1.
Investigating the Convergence of the Sequence \( aₙ = e^{-1/\sqrt{n}} \)
We are investigating the convergence of a sequence defined by the expression \( aₙ = e^{-1/\sqrt{n}} \). We’ll break down the process into clear steps.
1. Understanding the Expression
The expression \( aₙ = e^{-1/\sqrt{n}} \) represents the exponential function with a negative reciprocal of the square root of n in the exponent. As n increases, the exponent approaches 0, and we want to understand how this behaves as n approaches infinity.
2. Finding the Limit
To find the limit of the sequence as n approaches infinity, we can analyze the behavior of the exponent:
\( \lim_{n \to ∞} (-1/\sqrt{n}) = 0 \)
Since the exponential function is continuous, we can apply the limit to the entire expression:
\( \lim_{n \to ∞} e^{-1/\sqrt{n}} = e^0 = 1 \)
3. Conclusion
The sequence \( aₙ = e^{-1/\sqrt{n}} \) converges to 1 as n approaches infinity. This result provides insight into the behavior of the exponential function with a variable exponent that approaches 0. It illustrates how the exponential function behaves with complex expressions in the exponent and confirms the convergence of the sequence to 1.
Investigating the Convergence of the Sequence \( aₙ = \frac{{\ln(n)}}{{\ln(2n)}} \)
We are investigating the convergence of a sequence defined by the expression \( aₙ = \frac{{\ln(n)}}{{\ln(2n)}} \). We’ll break down the process into clear steps.
1. Understanding the Expression
The expression \( aₙ = \frac{{\ln(n)}}{{\ln(2n)}} \) represents the ratio of the natural logarithm of n to the natural logarithm of 2n. As n increases, both the numerator and denominator grow, and we want to understand how this behaves as n approaches infinity.
2. Finding the Limit
To find the limit of the sequence as n approaches infinity, we can use the properties of logarithms and simplify the expression:
\( aₙ = \frac{{\ln(n)}}{{\ln(2n)}} = \frac{{\ln(n)}}{{\ln(2) + \ln(n)}} \)
Now, we can divide both the numerator and denominator by \( \ln(n) \):
\( \frac{{\ln(n)/\ln(n)}}{{\ln(2)/\ln(n) + 1}} = \frac{1}{{\frac{{\ln(2)}}{{\ln(n)}} + 1}} \)
As n approaches infinity, the term \( \frac{{\ln(2)}}{{\ln(n)}} \) approaches 0, so the limit becomes:
\( \lim_{n \to ∞} \frac{1}{{\frac{{\ln(2)}}{{\ln(n)}} + 1}} = \frac{1}{0 + 1} = 1 \)
3. Conclusion
The sequence \( aₙ = \frac{{\ln(n)}}{{\ln(2n)}} \) converges to 1 as n approaches infinity. This result provides insight into the behavior of the ratio of logarithms of increasing numbers. It illustrates how the natural logarithm function behaves with complex expressions and confirms the convergence of the sequence to 1.
Investigating the Convergence of the Sequence aₙ = arctan(ln(n))
We are investigating the convergence of a sequence defined by the expression aₙ = arctan(ln(n)). We’ll break down the process into clear steps.
1. Understanding the Expression
The expression aₙ = arctan(ln(n)) represents the arctangent of the natural logarithm of n. As n increases, the natural logarithm of n grows, and we want to understand how the arctangent behaves as n approaches infinity.
2. Finding the Limit
To find the limit of the sequence as n approaches infinity, we can analyze the behavior of the natural logarithm of n:
lim (n → ∞) ln(n) = ∞
Now, we can apply the arctangent function to this limit:
lim (n → ∞) arctan(ln(n)) = arctan(∞)
Since the arctangent function has a horizontal asymptote at π/2, the limit becomes:
lim (n → ∞) arctan(ln(n)) = π/2
summary:lim (n → ∞) arctan(ln(n)) = arctan(lim (n → ∞) ln(n)) = arctan(∞) = π/2.
3. Conclusion
The sequence aₙ = arctan(ln(n)) converges to π/2 as n approaches infinity. This result provides insight into the behavior of the arctangent function applied to the natural logarithm of increasing numbers. It illustrates how the arctangent function behaves with complex expressions and confirms the convergence of the sequence to π/2.