Using the Limit Comparison Test In Exercises 17-26, use the Limit Comparison Test to determine the convergence or divergence of the series. 17. n/(n+1) 19. 1/(n^2+1) 21. 2n^2-1/(3n^5+2n+1) 23. 1/5^k+1 (k>2) 18. (n^2+1)/(4n+1) 20. (n+1)/(n^2+1) 22. n^2(n^2+4)/(n+1) 24. (n^2+1)/(n(n+1))^2n-1 26. 1/n
Added by Melvin C.
Close
Step 1
We can compare the series n/(n+1) with the series 1/n. Taking the limit as n approaches infinity, we have: lim(n->∞) n/(n+1) / (1/n) = lim(n->∞) n^2 / (n+1) = ∞ Since the limit is infinity, the series 1/n diverges. Therefore, by the Limit Comparison Test, the Show more…
Show all steps
Your feedback will help us improve your experience
Apratim De and 59 other Calculus 3 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Using the Limit Comparison Test In Exercises $17-26,$ use the Limit Comparison Test to determine the convergence or divergence of the series. $$\sum_{n=1}^{\infty} \frac{2 n^{2}-1}{3 n^{5}+2 n+1}$$
Infinite Series
Comparisons of Series
Using the Limit Comparison Test In Exercises $17-26,$ use the Limit Comparison Test to determine the convergence or divergence of the series. $$\sum_{n=1}^{\infty} \frac{2^{n}+1}{5^{n}+1}$$
Using the Limit Comparison Test In Exercises $17-26,$ use the Limit Comparison Test to determine the convergence or divergence of the series. $$\sum_{n=1}^{\infty} \frac{n}{(n+1) 2^{n-1}}$$
Recommended Textbooks
Calculus: Early Transcendentals
Thomas Calculus
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD