Solve the problem.
-If $\mathrm { p } 0$ and $\mathrm { q } 1$, what can be said about the convergence of $\sum _ { n = 2 } ^ { \infty } \frac { 1 } { { } _ { n P } ( \ln n ) ^q } ?$
A) Always diverges
B) May converge or diverge
C) Always converges

Question

Solve the problem.
-If $\mathrm { p } > 0$ and $\mathrm { q } > 1$, what can be said about the convergence of $\sum _ { n = 2 } ^ { \infty } \frac { 1 } { { } _ { n P } ( \ln n ) ^q } ?$
A) Always diverges
B) May converge or diverge
C) Always converges

Quizplus · Accepted Answer

We can use the integral test to determine the convergence of the series. Let $f(x) = \frac{1}{x^p(\ln x)^q}$. Then, $f(x)$ is positive, continuous, and decreasing for $x \geq e$, so we have
$$\int_2^\infty f(x) \, dx = \int_2^\infty \frac{1}{x^p(\ln x)^q} \, dx.$$
Making the substitution $u = \ln x$, $du = \frac{1}{x} dx$, we get
\begin{align*}
\int_2^\infty \frac{1}{x^p(\ln x)^q} \, dx &= \int_{\ln 2}^\infty \frac{1}{e^{pu} u^q} \, du \
&= \int_{\ln 2}^1 \frac{1}{e^{pu} u^q} \, du + \int_1^\infty \frac{1}{e^{pu} u^q} \, du.
\end{align*}
The first integral converges since $e^{pu} u^q \geq e^p$ for $u \leq 1$, and $\int_{\ln 2}^1 \frac{1}{u^q} \, du$ is a convergent $p$-series. The second integral converges by comparison to the integral $\int_1^\infty \frac{1}{u^q} \, du$, which is again a convergent $p$-series, so by the integral test, the series $\sum_{n=2}^\infty \frac{1}{n^p (\ln n)^q}$ may converge or diverge.

Solve the Problem p>0\mathrm { p } > 0p>0 And q>1\mathrm { q } > 1q>1

Solve the Problem $\mathrm { p } > 0$ And $\mathrm { q } > 1$