Can anyone help me start this problem from beginning to end, along with explanations on how to go about the problem for a better understanding how to do this series problem?
1) Find the values of p for which the series is convergent.
Summation notation symbol (with n=1 on bottom and infinity on top) of ln(n)/n^p
(Below, I try wording it in a mathematical sense, hoping it would format correctly here on the site).
∞
∑ ln(n) / (n^p)
n=1
see the solution here:
http://www.math.ucla.edu/~hendricks/CalculusIISummer2011/calc2practicefinalsolutions.pdf
Seeing other examples on the practice exam was really helpful.
Thank you Steve, it really helped. I will understand it a bit more better.
To determine the values of p for which the series is convergent, we need to use a convergence test. One commonly used test for the convergence of series is the integral test.
The integral test states that if the function f(n) is positive, continuous, and decreasing for n ≥ N (some positive integer N), then the series ∑ f(n) will have the same behavior as the improper integral ∫ f(x) dx from N to infinity. If the integral converges, so does the series, and if the integral diverges, so does the series.
In this case, the function f(n) is ln(n)/n^p. Since ln(n) and n^p are both positive for n ≥ 1, we can confirm that f(n) is positive for n ≥ 1.
Now, let's check if f(n) is decreasing for n ≥ N. To do this, we can take the derivative of f(n) with respect to n and analyze its behavior.
f'(n) = (1/n) - p(n^(p-1))/n^2 = (1/n) - p/n^p.
In order for f(n) to be decreasing for n ≥ 1, we need f'(n) to be negative for all n ≥ 1. Let's set f'(n) < 0 and solve for p:
(1/n) - p/n^p < 0
1 - pn/n^p < 0
1 < pn/n^p
n^p < pn
n^(p-1) < p
In order for this inequality to hold for all n ≥ 1, it must hold for n = 1. Substituting n = 1, we get:
1^(p-1) < p
1 < p
Therefore, the condition for f(n) to be decreasing for n ≥ 1 is p > 1.
Now, we can apply the integral test. We need to check the convergence of the integral ∫ f(x) dx from 1 to infinity.
∫ ln(x)/x^p dx = ∫ ln(x) x^(-p) dx
To solve this integral, we need to use integration by parts. Let's choose u = ln(x) and dv = x^(-p) dx.
du = (1/x) dx
v = (x^(1-p))/(1-p)
Applying the integration by parts formula:
∫ u dv = uv - ∫ v du
∫ ln(x) x^(-p) dx = ln(x) (x^(1-p))/(1-p) - ∫ (x^(1-p))/(1-p) (1/x) dx
= ln(x) (x^(1-p))/(1-p) - ∫ (x^(-p)) dx
= ln(x) (x^(1-p))/(1-p) + p/(p-1) x^(1-p)
Now, we can evaluate the integral from 1 to infinity:
∫[1 to ∞] ln(x)/x^p dx = [ln(x) (x^(1-p))/(1-p) + p/(p-1) x^(1-p)] from 1 to ∞
= lim(x→∞) [ln(x) (x^(1-p))/(1-p) + p/(p-1) x^(1-p)] - (ln(1) (1^(1-p))/(1-p) + p/(p-1) 1^(1-p))
Since ln(x) and x^(1-p) both grow without bound as x approaches infinity, the first term of the limit is infinite. Also, the second term evaluates to 0.
Therefore, the integral ∫[1 to ∞] ln(x)/x^p dx diverges when p > 1.
By the integral test, this implies that the series ∑ ln(n)/n^p also diverges for p > 1.
To summarize, the values of p for which the series ∑ ln(n)/n^p is convergent are p ≤ 1.