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Let $ \{ {b_n} \} $ be a sequence of positive num…

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Problem 41 Medium Difficulty

Let $ \{ {b_n} \} $ be a sequence of positive numbers that converges to $ \frac {1}{2}. $ Determine whether the given series is absolutely convergent.

$ \displaystyle \sum_{n = 1}^{\infty} \frac {{{b_{n }^{n} \cos n \pi }}}{n} $

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Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 11

Infinite Sequences and Series

Section 6

Absolute Convergence and the Ratio and Root Tests

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Video Transcript

What bnb A sequence, A positive numbers conversion toe. One half. We want to know whether the Siri's convergence so as usual, let's denote this term and turn by an and let's try to apply the root test and we'LL see why I went to the Rue test of one second First noticed that co sign and pie. If you start plugging in values of end equal to one, two and so on, you see that the sequence starts off a negative one one and so on. So the sequence is really equal to negative wants to the end. So that means that my end, it's beyond to the end with negative once in the end. So here and the root tests, we'd like to look at the end through. So let me write this. You take your A m absolute value of that and then raise that to the one over an hour and I take the limit as n goes to infinity. This is the route test. Now we're taking up slew value, So that means we could drop the negative one. And when we raise B end to the end to the one over end, recall that you take those two exponents, multiply them together. In this case, that's just one. And here we don't need absolute value for being because we're already told that they're positive. So now my limit becomes I have just being on the top and now on the bottom here we'LL just have and one over. And so now it suffices to consider this last limit. You might have maybe memorized the formula involving this, but here I can show some work. So the problem with the limit in the denominator is that it's an indeterminate form of the type infinity to the zero power. So a nice way to deal with these is to do the following trick. So I'll use the fact that any number X could always be written as e to the natural log of X. So here I'LL do this formula using this is my ex value. So have e natural log of my ex and then here use your log properties to bring the one over end outside of the log and then now the limit is in the exponents. This is infinity over infinity. So you would go ahead and use Lopez House rule here so the derivative of natural log is one over end in the derivative of end is just one. So now this limit is just e to the zero, which equals one. So that means that the denominator approaches one so that this limit over here, where we stopped earlier, this limit is now just limited bm which is a half over this limit, which we just computed, which was one giving us an answer. What half? Now that's less than one. So by the root test, the Siri's is absolutely conversion. Yeah.

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