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

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

$p>0$

Calculus 2 / BC

Chapter 11

Infinite Sequences and Series

Section 5

Alternating Series

Sequences

Series

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let's find out for what value of P. The series is conversion, so let's look in some cases here. So case one. Let's say he's negative. Then here. If he is negative, then we can rewrite a n The answer So into the minus p. This will while appears negative then negative ps Baber than zero. So as we take and to go into infinity, this will be infinity. And then when we multiply it by this negative one, this tells us that a n ossa lates between plus and minus infinity. So, in other words, in political or this implies the limit of an is down zero. So the sun diverges in this case. So it's right that over here diverges by the test four diversions. That's what your book calls it. So we're not going to include this case in our answers, So that's not a part of the solution. So now we'LL go to the next beach. Let's say P equals zero. It's okay. Still, then a n is just negative one to the end, minus one into the zero. So just minus wants you to invite us one and therefore the limited and does not exist in particular. It's not equal to zero because it just oscillates between that one and one. So once again, the Siri's will divert by the diversions test, and we have another case to consider. Case three Coming up shortly. So now, Now let's go on to the next page for Case three Soap. You argue that zero. So let's look at our end. And then here. Let's define being end to just be one over into the peak. And let's try the altering Siri's test so we could see that Vienna's bakers and zero positive divided by a positive. And we have our other conditions to satisfy. We know that we must have that the limit of being zero that will people zero because and this positive so the denominator will go to infinity, and we need that beyond. The sequence is decreasing if this is true that this original series will converge when Pius positive, so this needs to be checked. So the way to show this is to you, Khun, just look at the derivative of dysfunction. It's called X minus p. So F crime of X minus P x to the mind his P minus one as using the powerful, and this is negative sense. He negative p is less than zero. But here X is bigger than you know, because Axe is taking on the same values of end. So X has to be bigger than one just like and wass So the original series, well converge in this case converges if and only if he is bigger than zero, and that's our final answer.

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