%5En-ln(n)%2Fsqrt(n)-convergence.jpeg "(-1)^n Ln(n)/sqrt(n) Convergence")
Investigating the Convergence of (-1)^n ln(n)/sqrt(n)
This article explores the convergence of the sequence (-1)^n ln(n)/sqrt(n). We will utilize the Alternating Series Test to determine if the series converges or diverges.
The Alternating Series Test
The Alternating Series Test states that an alternating series of the form:
∑ (-1)^n * a_n
converges if:
- a_n > 0 for all n
- a_n ≥ a_{n+1} for all n
- lim_{n→∞} a_n = 0
Applying the Test to (-1)^n ln(n)/sqrt(n)
Let's examine our sequence in light of the Alternating Series Test:
-
a_n = ln(n)/sqrt(n). Since the natural logarithm of a number greater than 1 is always positive, and the square root of a positive number is also positive, a_n > 0 for all n > 1.
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To verify a_n ≥ a_{n+1}, we need to analyze the function f(x) = ln(x)/sqrt(x) and show it's decreasing for x > 1. Taking the derivative of f(x) using the quotient rule, we get:
f'(x) = (sqrt(x) - ln(x))/(2x*sqrt(x))
For x > 1, the numerator is negative (since ln(x) > 1 for x > e) and the denominator is positive. Therefore, f'(x) < 0, confirming that f(x) is decreasing, and consequently, a_n ≥ a_{n+1} for all n > 1.
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lim_{n→∞} ln(n)/sqrt(n) = 0. This can be proven using L'Hopital's rule, as both the numerator and denominator approach infinity as n approaches infinity. After applying L'Hopital's rule, we get:
lim_{n→∞} (1/n)/(1/(2sqrt(n))) = lim_{n→∞} 2sqrt(n)/n = lim_{n→∞} 2/sqrt(n) = 0
Since all three conditions of the Alternating Series Test are satisfied, we can conclude that the series ∑ (-1)^n ln(n)/sqrt(n) converges.
Conclusion
Using the Alternating Series Test, we have demonstrated that the sequence (-1)^n ln(n)/sqrt(n) converges. This analysis highlights the importance of carefully examining the behavior of a sequence to determine its convergence properties.