Question

In Exercises 67–72, use the results of Exercises 63 and 64 to determine if each series converges or diverges.∑(from n=2 to ∞) [(ln n)¹⁰⁰⁰ / n¹.⁰⁰¹]

Accepted Answer

Identify the general term of the series: $$a_n = \frac{(\ln n$$)^{1000}$$}{n$$^{1.001}$$}$$ for $$n \geq 2. $$Recall the p-series test: a series of the form $$\sum \frac{1}{n^p}$$ converges if and only if $$p \u003e 1. $$Here, the denominator has $$n^{1.001}$$, which suggests a p-series with $$p = 1.001 \u003e 1. $$Consider the numerator $$(\ln n$$)^{1000}$$, which grows slower than any positive power of n$. This means the logarithmic term grows slower than any polynomial term in the denominator. Use the Comparison Test or Limit Comparison Test by comparing a_n$ with b_n$$ = \frac{1}{$$n^{1.001}$$}$$. Since $$\lim_{n 	o \infty} \frac{a_n}{b_n} = \lim_{n 	o \infty} (\ln $$n)^{1000}$$, which tends to infinity, the direct comparison test is inconclusive, so use the Limit Comparison Test carefully. Apply the Limit Comparison Test with $$b_n = \frac{1}{n$$^{1.001}$$}. $$Compute $$\lim_{n 	o \infty} \frac{a_n}{b_n} = \lim_{n 	o \infty} (\ln n$$)^{1000}$$. Since this limit is infinite, the Limit Comparison Test suggests that a_n$ behaves like a larger term than b_n$, but since b_n$ converges, check if the logarithmic growth affects convergence by using the Integral Test or Cauchy Condensation Test.

In Exercises 67–72, use the results of Exercises 63 and 64 to determine if each series converges or diverges.
∑(from n=2 to ∞) [(ln n)¹⁰⁰⁰ / n¹.⁰⁰¹]

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Comparison Test for Series

Behavior of p-Series

Growth Rates of Logarithmic vs. Polynomial Functions