Question: A). Use the Limit Comparison Test to determine whether the series converges or diverges. ∞ n = 1 1 2(1 + 1/n)n(1 + 1/n) Identify bn in the following limit.

Using the limit comparison test, test if the series is convergent or divergent. infinity n = 1 n/(n + 1)2n ...

Testing for Convergence In Exercises 35-64, use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer. 41. ∫ 0 π t + s i n t d t 42. ∫ 0 t t − s i n t d t (Hint: t ≥ sin t for t ≥ 0) 43. ∫ 0 2 1 − x 2 d x 44. ∫ ...

Use the limit comparison test to determine whether each of the following series converges or diverges. 211. 4"-3" n=1 Use the limit comparison test to determine whether each of the following series converges or diverges. 215. Σ ,1 + 1/n n=in

Use the Limit Comparison Test to determine the convergence or divergence of the series. 00 1 n n = 1 'n6 + 4 1 n (16+4 lim n-> 1 = L > 0 x n 6 converges diverges Use the Limit Comparison Test to determine the convergence or divergence of the series. E n2n+ 5) n2(n + 5) n = 1 1 n2(n + 5) 1 lim n --> 00 = L> 0 n 3 converges diverges Use the Limit

Math; Calculus; Calculus questions and answers; Use the Limit Comparison Test to determine whether the series converges or diverges.∑n=1∞an=∑n=1∞9*12n+6*8n13n+4nThe comparison series is ∑n=1∞bn=∑n=1∞arn-1 where a= and r= Then, limn→∞anbn= ∑n=1∞bn is a , therefore ∑n=1∞an by the Limit Comparison Test.

Question: Use the limit comparison test to determine if the following series converges or diverges Choose the correct answer below. 0 A. The limit comparison test with an- and bshows that the series converges. 5n O B. The limit comparison test with an- nd bn shows that the series diverges. n 4 O C.

Use Limit comparison test \sum^{\infty}_{n=1} (\sqrt [n] {e}-1) Use the Limit Comparison Test. \sum_{n=1}^{\infty} \frac{1}{\sqrt[3]{2n^2+5 \sum_{n = 1}^{\infty} \frac{1}{\sqrt{n^3 + 1 Use The Limit Comparison Test. Use the comparison test or the limit comparison test to evaluate \sum_{k=1}^{\infty} \dfrac{2 + \sin k}{k^3}. What is the ...

Explain how the Limit Comparison Test works. Choose the correct answer below. O A. Find an appropriate comparison series. Then determine whether the terms of the given series are less than or equal to or greater than or equal to for all large values of k. This comparison determines whether the series converges. OB. Find an appropriate ...

- Use the Limit Comparison Test to determine the convergence or divergence of the series. Be sure to identify the comparison series and how you know whether it converges of diverges before beginning the Limit Comparison Test. ∑ n = 1 ∞ n 2 (n 2 + 4) 1