Solve the problem.
-If $\mathrm { p } 1$ 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 $\mathrm { p } > 1$ 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

Since $p>1$ and $q>1$, we have $n_p>(\log n)^q$ for sufficiently large $n$ (by L'Hopital's Rule, for example). Thus, $\frac{1}{n_p(\log n)^q}<\frac{1}{(\log n)^{2q}}$ for sufficiently large $n$, and $\sum_{n=2}^\infty\frac{1}{(\log n)^{2q}}$ converges by the integral test. Therefore, by the comparison test, $\sum_{n=2}^\infty\frac{1}{n_p(\log n)^q}$ converges.

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

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