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Show that series $ \sum ( - 1)^{n-1}b_n, $ where $ b_n = 1/n $ if $ n $ is odd and $ b_n = 1/n^2 $ if $ n $ is even, is divergent. Why does the Alternating Series Test not apply?

The Alternating Series Test does not apply since $\left\{b_{n}\right\}$ is not decreasing.

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Let's show that this Siri's diverges and then we'LL explain why we're not allowed to use the alternating, Siri says for this problem. So let's just write this. Siri's out. Let's say it's starting from one to infinity. Negative one. The starting point doesn't matter. All that really matters is the pattern. So here, when we plug in and equals one, you have B one, So that's one over one. So this should be be one minus B two plus B three minus before and so on. So that's one over one minus one over two square. So using that the formula used depends on weather and is even Rod One is odd. So you used one over end who is even so one over and swear three desired and so on. Now we can rewrite this so notice that let's just combine the positive terms and then separately combined the negative terms. So she's pause that minus, and then we have one over two squares, one over force where one over six weird and so on. Now we can rewrite this and the signal station. That's the first Siri's, and then for the second one. He's a different color here, minus the sum from n equals, one to infinity of one. Over there we have, but still to win and then square. Now this second, Siri's and blue. This Siri's does converge, so you can just use the pee test here with people's too, which is bigger than one as the pee test from section eleven point three. However, this first series we can show that the Siri's diverges. So just show that this diverges. You can use the Lim comparison test with B and E equals one over end. So call this your a n use limit comparison. You know that the sum of the d n diverges this's infinity harmonic series. Therefore, this Siri's also will diverge by the limit comparison test. So you have a diversion. Siri's minus this really number because the second one did converge. So let me write that in the next page you have diversion minus conversion. Well, this is just a real number. So if you have in our case, infinity and you subtract just a number, that doesn't really do anything to infinity, so the result will be diversion. And that's what we wanted to show for the first part The second part of the question was why couldn't we use a S T alternating series test? That's what I mean by S T. Why can't this d applies? And this problem And the reason was because of the way the BM was written so bien was equal to one over end if and was are so piece wise function. But it was one over and swear, if n is even and you can show that this function is is not decreasing. So, for example, if we look at like, say, a baby too is one over four. But then B three equals one over three. So from there we increased. But then when we go back to before, that's one over sixteen. So if this is not increasing anymore, so in general we see that if if n is on, then Bien will be bigger than so here it will be bigger than the previous value. But then we have d n minus one is less than being minus two. So the problem here is that if you're going from even and odd that you increase But then when you go from ombati, even you decrease So this see this sequence is neither increasing nor decreasing. This is why we're not allowed to use the altar in seriousness, and that's your final answer.