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JH
Numerade Educator

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Problem 41 Hard Difficulty

Determine whether the series is convergent or divergent. If it is convergent, find its sum.
$ \displaystyle \sum_{n = 1}^{\infty} \left( \frac {1}{e^n} + \frac {1}{n(n + 1)} \right) $

Answer

Convergent, $\frac{e}{e-1}$

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Video Transcript

let's determine whether the given Siri's converges or diverges, and if it does converge, then we'LL go ahead and find the sun. So instead of looking at this entire sung, let's just consider these smaller Siri's over here and that these both converge individually. Then we'LL just Adam phone together, and that will be the sun. So this Siri's right here. Let's re write it because it's actually geometric. It's more obvious to see that if you write it in this form over here and we see that this is geometric our equals one over easy and that satisfies this inequality here. So in our problem, that's Lou value our peoples want to relieve. That's less than one, definitely because one is less than me. So that means that this geometric series will converge and we could even say what the sum is. This is again. This is all for the series here. So for geometric, we know that the sum equals you. Look at the first term of the series, and then you do one minus R. So in this case, the first term, that's when you plug in and equals one here. So if you plug that in that first term is just one overeat to the first power and then one minus one over easy. And this could be simplified to one over E minus one. So this is the sum of the geometric series. Now we'LL go on to the other series remaining over here. If this ends up converging, will find the sum And then we'LL add this green Cem to this term over here and I don't be the sum of the entire series. So I'm running out of room here. Let me go on to the next page to deal with the second term over here. So for something like this, we should take this term over here. If this is not geometric, we should do partial fraction the composition. So we'LL have to find values of a plus B will satisfy this And when we do that, we should end up with one over end. So is one. And then B is minus one. Also, let me go ahead and rewrite. This is a limit. So instead of taking the whole infinite some I'm just taking the limit of the cave partials on and then we'LL do the telescoping Siri's That's the type of Siri's that this is This's what the textbook calls in telescope E So we won't evaluate the limit right now we'Ll just keep writing limit And then this is where we just simplify this the whole summation right here. So let's just start at plugging in n equals one. We just have one minus a half and then plug it and equals two and equal three and then and so on. And then you were at the very end you were plugging K minus one and then right after that, the very last term you plug in K. So let me go ahead and indicated with a different color here. So this term down here this was corresponding to an equals one and equals two and equals three. And then we skipped a few because we can write them all and then we come over here. This is K minus one. The second the last guy And in the very last one over here was K. Now we do the telescoping. This is when we cancels much as we can. We could cancel the half one third one fourth and we would keep canceling all the way until we get to one of her. Okay, but then we're still left over with this minus. So what are we loved over with? We have this one, and then we have this. So this is limited. Kay goes to infinity of one minus one over Kaye plus one. So now let Kay go to infinity. And what will happen is this fraction will go to zero, and we're just left over with one. So now for our final answer. Well, let's just write this out on the next page. Our original theories What? This is equal to just the sum of the two series. And the reason I know this is true is because we just sure that both of these two Siri's converge here and we actually found the sum. And when we found the sum for the first series, that was just one over e minus one. That was in the first page, and we did that in blue and then on the page, too. We did the telescoping and just found the sum over here. That's one. And of course, if you want that common denominator, just go ahead and multiply this one top and bottom by minus one sea of one plus you minus one overeem minus one. Celeste cious e over e minus one. So the final answer is that given Siri's does converge and the sum of the entire Siri's is e divided by minus formed, and that's our final answer.