Using the integral test, find the positive values

Answered question

2022-04-02

Using the integral test, find the positive values of p for which the series ∑∞ k=2 1/ (k(ln(k)))^p converges. Show your work and explain your reasoning.

Answer & Explanation

alenahelenash

alenahelenash

Expert2023-04-27Added 556 answers

We are given the series k=21(kln(k))p, and we want to find the values of p for which this series converges. We can use the integral test to do so.
According to the integral test, if nf(x)dx converges, then k=nf(k) converges, and if nf(x)dx diverges, then k=nf(k) diverges. Thus, we need to determine the convergence of the integral 21(xln(x))pdx.
We can simplify this integral using a substitution u=ln(x), which gives us:
21(xln(x))pdx=ln(2)1(ueu)peudu=ln(2)e(1p)uupdu
For this integral to converge, we need (1p)u<0 as u (since u=ln(x)). Therefore, we must have p>1.
Now, let's consider the case where p>1. Then, we can use the limit comparison test with the series k=21kp, which we know converges if p>1. Specifically, we have:
limk1(kln(k))p1kp=limkkp(kln(k))p=limk(1ln(k))p=0
Since the limit is finite and positive, and since k=21kp converges if p>1, it follows that k=21(kln(k))p also converges if p>1.
Therefore, the series k=21(kln(k))p converges if and only if p>1.

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