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# Determine whether the sequence converges or diverges. If it converges, find the limit.$a_n = \frac { (\ln n)^2}{n}$

## Converges to 0

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for this problem when we're looking at limit as n goes to infinity of a n. Wendell to use low Patel's rule here. If you want to just plug in and equals infinity, then we'd have infinity on top and Cindy in the denominator as well. So were, of course, not allowed to just plug in infinity because infinity over INF India's indeterminate form. But if you are in the situation where you would get infinity over and Cindy, then low Patel's rule is applicable. Hello, petals. Rule is you know, when you're in this infinity over infinity or zero over zero type of setup. And when you're in that type of situation, you're allowed to just do the derivative of the numerator and divide by the derivative of the denominator. So the derivative of the numerator is too time's natural log of n to the two minus one, which is one times the derivative of the inside function. So times one over in through than the chain rule to this guy, and then we're dividing by the derivative of in with respect to end. So we're just dividing by one, and now we can rewrite this as limit as n goes to infinity of two times natural log of n divided by n and again. We're in a situation where we can use low Patel's rule. So this is limit as n goes to infinity of two times one over end divided by one. And this limit goes to zero. So our sequence does converge and it converges to zero.

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