Find the values of p for which the following series is convergent.
The sum from n=2 to infinity of 1/(n(ln(n))^p).
I tried the integral test using u-substitution, and I know the answer is p>1, but I'm not quite sure how to get there. Thanks for the help.
To determine the values of p for which the given series is convergent, let's use the integral test.
First, we need to integrate the function f(x) = 1/(x(ln(x))^p). To apply the integral test, we need to integrate from 2 to infinity since the given series starts from n = 2.
Let's perform the integral:
∫(1/(x(ln(x))^p)) dx
To integrate this function, we can use the substitution method. Let u = ln(x), then du = (1/x) dx. Rewriting the integral using u:
∫(1/(x(ln(x))^p)) dx = ∫(1/(u^p)) du.
Now, we can integrate the function with respect to u:
∫(1/(u^p)) du = (u^(1-p))/(1-p) + C,
where C is the constant of integration.
Substituting u back in terms of x:
(x^(1-p))/(1-p) + C.
Next, we will evaluate the definite integral from 2 to infinity:
∫ from 2 to infinity (1/(x(ln(x))^p)) dx = [(x^(1-p))/(1-p)]| from 2 to infinity.
To check the convergence, we consider the limit of this expression as the upper limit goes to infinity:
lim from x =∞ [(x^(1-p))/(1-p)] - [(2^(1-p))/(1-p)].
If this limit exists (finite), then the series converges. Otherwise, if the limit is infinite or undefined, the series diverges.
To simplify the limit, we need to consider some cases for p.
Case 1: p > 1
In this case, when x goes to infinity, the term x^(1-p) goes to zero. Thus, the limit becomes:
lim from x =∞ [(x^(1-p))/(1-p)] - [(2^(1-p))/(1-p)] = [(2^(1-p))/(p-1)].
Since this limit is finite, the series converges for p > 1.
Case 2: p ≤ 1
In this case, when x goes to infinity, the term x^(1-p) goes to either infinity or it diverges. Therefore, the limit is infinite or undefined.
Therefore, the series only converges for p > 1.
Hence, we have found that the values of p for which the given series is convergent are p > 1.