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JH
Numerade Educator

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Problem 63 Hard Difficulty

Find the values of $ x $ for which the series converges. Find the sum of the series for those values of $ x. $
$ \displaystyle \sum_{n = 0}^{\infty} e ^{nx} $

Answer

$$\frac{1}{1-e^{x}}$$

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Video Transcript

Let's find the values of extra wish, the Siri's comm urges. So that will be one part of this problem here. Let's call that part one. And then for those values of X in part one. Well, go ahead and find the sum So that'LL be the second part of this problem. So for the first part here, let's just rewrite the Siri's. It's actually geometric, but it's it might not be obvious by the way, that it's Rin. So here I would just right This is either the eggs to the end power here. Understood. Using your laws of exponents you could always rewrite. This is a B times C. So if you want, you could always pull out one of the exponents here and I just chose to pull up the end. So now I can see that this is geometric and R r is equal Teo either the ex So once excess fixed, this is your r value and we know that geometric that these will converge, converge if we have the absolute value are is less than one. So in this case, this will converge. If the absolute value of either the ex was just either the access is always positive, so the absolute value is always equal to itself. And we need this to be less than one. So now we have an inequality here that we need to solve for X because in part one, we do want the ex values. So let's take the natural log of both sides here. So this's since the log function is increasing, the bigger the importance than the bigger the log will be. So since one is bigger than you, fax Ellen of one will have to be bigger than Ellen either. The ex. Yeah, so here on the left side, using the fact that Ellen and E. T. The extra inverse is this is just X and Ellen of one zero or so That's our answered apart one. We want X to be a negative number, and then the series will converge. Now we go to Part two, where we actually find the sum of the Siri's. Assuming that, of course, except a negative number. So it's gonna part two. Now let's hasten room here. So for part two for geometric series, we know that there's some equals. You take the first term of the series over one minus the are the common ratio. So in this problem, the first term happens when you plug in and equal zero. They're so plugged that in, friend and you get a one, you get E X. But that's all being raised with zero power. So you just get one and then one minus R, which is just east of the ex. That's from our are Step One, and that's our final answer. And of course, the sum only is true when we're assuming that exes less than zero, because that's when the series converges, So that's our final answer.