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For what value of $ p $ is each series convergent?

$$ \displaystyle \sum_{n = 1}^{\infty} \frac {( - 1)^n}{n + p} $$

$p$ can be any real number

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Campbell University

Oregon State University

University of Nottingham

let's determine for which values of P the Siri's his conversion. Well, first of all, we know that end will take on the value, staring at one, then two, three, and so on. So these air natural numbers. So we do not want Peter to be one of these numbers, and we're the negative of one of peeps. So first of all we want and plus P to be non zero because it's in the denominator. So this means we want pizza and not be negative end. So we do not want the negative of these numbers and so on. So we can also write this more formally. So we claim that we just want Peter's just be of not of the form and where and is an international number. So the sorry I should have been a put a negative improvement that end not equal to negative and where and is natural otherwise go. The Siri's is under fine cause. We have zero ended denominator in one of the terms, and that's not allowed. So first of all otherwise now, well, suppose that he is now one of these numbers and then we'll show that it actually convergence so now suppose he is not equal toe negative end for any of these naturals. Then let's define B end to be the absolute value of this. Answer him over here. So if we take the absolute value that the numerator, those one but the denominator can be negative depending on P piece negative number, the denominator could end up being negative. So we have absolute value there now, the first condition here. So what we're using now is the alternating Siri's test, and the reason it's good to use this test or at least try to use it is because we have this all streaming series. So the first condition is that the bee ends or bigger than or equal to zero. And that's true. We have a positive divided by a positive, so the fraction is positive. So check the second condition. Is that the limit as N goes to infinity of being a zero, and this will be true because he is just fixed so eventually the denominator will get larger, larger and even that you could even drop the absolute values eventually because in the limit and will be larger than negative P. So you could even drop those if you want to the denominators going to infinity, the liberators just want so that limit is zero, and there's one more condition to check. So I'LL put that on the next page. So we have Bee n equals one over Absolute Value and plus P. And let's just not that this is equal to one over and plus P if and plus P is bigger than zero. So in other words, and is bigger than negative p. So let's suppose and is bigger than negative p. It's eventually bigger than I gotta be because natural number and we're starting at one going all the way to infinity. So at some point, this inequality will eventually be true. So this is just not there on the side. Eventually, true, as an increases now we want to show that this sequence is decreasing. That's equivalent to showing the following term is less than or equal to the previous term. So this is true. This is equivalent to and plus one plus p being less than or equal to one over and plus P. And this is a true inequality here because these numbers are both positive recall and plus piece. Where is bigger than zero? This is our assumptions. These are both positive. But the denominator on the left is larger. Yeah, no, it's not just that. It's larger X that also they're both positive. Both denominators, our positive bigger than zero. Therefore, we've just shown that this being is decreasing. Now we can apply the alteration. Siri's test. We've checked all the conditions. The bien was positive. Check the limit of the B N was zero check and then be in decreasing. Check. So bye, the alternating Siri's dirham the Siri's convergence, and that's our final answer.