Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Consider the series sigma_n = 1^infinity (3n - 1/6n + 3)^2n. Evaluate the the fo

ID: 2861368 • Letter: C

Question

Consider the series sigma_n = 1^infinity (3n - 1/6n + 3)^2n. Evaluate the the following limit. If it is infinite, type "infinity" or "inf'. If it does not exist, type "DNE". lim_ n rightarrow infinity^n squareroot |a_n| = L What can you say about the series using the squareroot Test? Answer "Convergent", "Divergent", or "Inconclusive". Determine whether the series is absolutely convergent, conditionally convergent, or divergent. Answer "Absolutely Convergent", "Conditionally Convergent", or "Divergent".

Explanation / Answer

lt (3n-1)^2/(6n+3)^2=3^2/6^2=9/36

limit is less than 1 so series is convergent.

since mode of limit is also less than 1, series is absolutely convergent.

Related Questions

Consider the series RLC circuit shown in Figure 1 The switch has been open for a
Question #1810354
Consider the series RLC circuit shown in Figure 1 The switch has been open for a
Question #1938763
Consider the series an where an = (n+4)!/en-7 In this problem you must attempt t
Question #3193603
Consider the series an where an = en+8rootn+4 / (n + 6)! In this problem you mus
Question #3082239
Consider the series infinity n=1 (-1)n+1 8n-1/(n+4)6 7n+2 In this problem you mu
Question #3192587
Consider the series infinity n=1 14n/(n=1)102n=1 In this problem you must attemp
Question #3192585
Consider the series infinity n=1 an where an= (-10)n+5/(2n + 5)12n In this probl
Question #3192583
Consider the series infinity n=1 an where an= en-2/(n=4)! In this problem you mu
Question #3192584

Navigate

Previous
Consider the series sigma^infinity_n = 1 a_n where a_n = (-1)^n n/n^2 + 5n +3 In
Next
Consider the series sigma_n=1^inifinty a_n where a_n = (n+3)!/e^n+9 squareroot n