Question

Loglog p-series Consider the series ∑ (k = 2 to ∞) 1 / (k(ln k)(ln ln k)ᵖ), where p is a real number.
b. Which of the following series converges faster? Explain.
∑ (k = 2 to ∞) 1 / (k(ln k)²) or ∑ (k = 3 to ∞) 1 / (k(ln k)(ln ln k)²)?

Accepted Answer

Step 1: Understand the two series given for comparison. The first series is $$ \sum_{k=2}^\infty \frac{1}{k (\ln k)^2} $$ and the second series is $$ \sum_{k=3}^\infty \frac{1}{k (\ln k) (\ln \ln k)^2} . $$Both are infinite series with terms involving logarithmic functions in the denominator. Step 2: Recall that the speed of convergence of a series depends on how quickly its terms approach zero. Smaller terms generally mean faster convergence. So, we want to compare the size of the terms $$ \frac{1}{k (\ln k)^2} $$ and $$ \frac{1}{k (\ln k) (\ln \ln k)^2} $$ for large $$ k . $$Step 3: For large $$ k $$, note that $$ \ln k \u003e 0 $$ and $$ \ln \ln k \u003e 0 . $$Since $$ (\ln k)^2 $$ grows faster than $$ (\ln k)(\ln \ln k)^2 $$ because $$ (\ln \ln k)^2 $$ grows slower than $$ \ln k $$, the denominator $$ k (\ln k)^2  is $$larger than $$ k (\ln k)(\ln \ln k)^2 $$ for sufficiently large $$ k . $$Step 4: Since the denominator of the first series' terms is larger, its terms are smaller compared to the second series' terms for large $$ k . $$Smaller terms imply that the first series converges faster than the second series. Step 5: To confirm this rigorously, one could apply the Integral Test or compare the series using the Limit Comparison Test by examining the limit of the ratio of their terms as $$ k 	o \infty . $$This will show which series' terms decrease faster, confirming the conclusion about convergence speed.

Loglog p-series Consider the series ∑ (k = 2 to ∞) 1 / (k(ln k)(ln ln k)ᵖ), where p is a real number.
b. Which of the following series converges faster? Explain.
∑ (k = 2 to ∞) 1 / (k(ln k)²) or ∑ (k = 3 to ∞) 1 / (k(ln k)(ln ln k)²)?

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Comparison Test for Series Convergence

Logarithmic and Log-Logarithmic Series

Rate of Convergence

Loglog p-series Consider the series ∑ (k = 2 to ∞) 1 / (k(ln k)(ln ln k)ᵖ), where p is a real number.b. Which of the following series converges faster? Explain.∑ (k = 2 to ∞) 1 / (k(ln k)²) or ∑ (k = 3 to ∞) 1 / (k(ln k)(ln ln k)²)?

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Comparison Test for Series Convergence

Logarithmic and Log-Logarithmic Series

Rate of Convergence

Loglog p-series Consider the series ∑ (k = 2 to ∞) 1 / (k(ln k)(ln ln k)ᵖ), where p is a real number.
b. Which of the following series converges faster? Explain.
∑ (k = 2 to ∞) 1 / (k(ln k)²) or ∑ (k = 3 to ∞) 1 / (k(ln k)(ln ln k)²)?