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.

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alenahelenash · Accepted Answer

We are given the series ∑k=2∞1(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 ∫n∞f(x)dx converges, then ∑k=n∞f(k) converges, and if ∫n∞f(x)dx diverges, then ∑k=n∞f(k) diverges. Thus, we need to determine the convergence of the integral ∫2∞1(xln(x))pdx.We can simplify this integral using a substitution u=ln(x), which gives us:∫2∞1(xln(x))pdx=∫ln(2)∞1(ueu)peudu=∫ln(2)∞e(1−p)uupduFor this integral to converge, we need (1−p)u&amp;lt;0 as u→∞ (since u=ln(x)). Therefore, we must have p&amp;gt;1.Now, let&#039;s consider the case where p&amp;gt;1. Then, we can use the limit comparison test with the series ∑k=2∞1kp, which we know converges if p&amp;gt;1. Specifically, we have:limk→∞1(kln(k))p1kp=limk→∞kp(kln(k))p=limk→∞(1ln(k))p=0Since the limit is finite and positive, and since ∑k=2∞1kp converges if p&amp;gt;1, it follows that ∑k=2∞1(kln(k))p also converges if p&amp;gt;1.Therefore, the series ∑k=2∞1(kln(k))p converges if and only if p&amp;gt;1.

Using the integral test, find the positive values

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