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

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.

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