Find limit of each sequence or state DNE. 1. $a_n = \frac{\ln(n^2+1)}{n}$ 2. $a_n = (-1)^{n+1}\frac{2n+1}{3n-1}$
Added by Cindy C.
Close
Step 1
Step 1: For the first sequence, we have $$ \lim_{n\to\infty} a_n = \lim_{n\to\infty} \frac{\ln(n^2+1)}{n} $$ This is of the indeterminate form $\frac{\infty}{\infty}$, so we can use L'Hopital's rule: $$ \lim_{n\to\infty} \frac{\ln(n^2+1)}{n} = \lim_{n\to\infty} Show more…
Show all steps
Your feedback will help us improve your experience
Madhur L and 98 other Calculus 1 / AB educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Problem #1: State whether the sequence converges or diverges. If the sequence converges, find its limit: A = ln(n) * (-1)^(n+1) * ln(n^3), n=3 B = 2 * (-1)^(7+11+37), n=3 Problem #2: Determine whether the given series Converge Absolutely, Converge Conditionally, or Diverge and give reasons for your conclusions: (-1)^(7+13+n^7), n=3
Madhur L.
Use Theorem 1 to determine the limit of the sequence or state that the sequence diverges. $$ a_{n}=\ln \left(\frac{12 n+2}{-9+4 n}\right) $$
Adi S.
Consider the sequence with terms an = (2n / (2n - 7))^n (n = 1, 2, 3, ...) Determine whether {an} converges or diverges. If the sequence converges, find its limit. Do the following series converge or diverge? Justify your answer. (i) sum n=1 to infinity (n^2 e^n) / n! (ii) sum n=1 to infinity (3n + 1) / (2n^2 + n cos^2 n + 1)
Vincenzo Z.
Recommended Textbooks
Calculus: Early Transcendentals
Thomas Calculus
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD