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JH
Numerade Educator

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Problem 47 Medium Difficulty

Determine whether the series is convergent or divergent by expressing $ s_n $ as a telescoping sum (as in Examples 8). If it is convergent, find its sum.
$ \displaystyle \sum_{n = 1}^{\infty} \left( e^{1/n} - e^{1/(n + 1)} \right) $

Answer

$e-1 .$ Converges

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Video Transcript

Let's determine whether the Siri's converges air diversions by expressing SN as a telescoping some. And then, if it converges, will go and find that summer's well. So, SN recalled, by definition is just the sum of the first and numbers where your a N is this thing that's being added up. So let's just go ahead and write. This is a limit. Kay goes to infinity, and then we have the sum from n equals one. Take a hee won over in minus E one over and plus one. Now this term here in the apprentices, this whole sum this is S K. That's the sum from one all the way of Tien, she says. It's defined. So let's telescope. Let's do a telescoping, some on ly inside of the prophecies. And then after that, we'LL come out here and take the limit, and that should answer our question. So let's write this. So now let's go ahead and start adding, so plug in and equals one first. That's our starting point. So eat to the one over one minus E to the one over two. This's the first term after plugging in and equals one and then next you increase and buy one. So and is now too. It's an equals two there, maybe one more term in this direction. So hee So now industry. So plug that in and you might see some cancellation happening there already, so we would keep going in this direction. So put some dots there to indicate that. And then eventually, at the very end, when we plug in and equals care would have ete to the one over K minus. Eat the one over Kaye plus one. But in order, Tio, maybe help you see the pattern here, it might be best to write a term before n equals K. So the term right before the last one came in this one and by writing that you could see some more cancellation over here, the e to the one over K will cancel with negative eats of the one of Okay, so now let's keep see how much we could cancel each of the one over one. This doesn't cancel with anything because all the remaining denominators are larger than one. Here we have negative eats of the one half that'LL cancel with the positive each other one half more cancelation there. If I wanna keep writing, I would see that e to the neck. Negative. Each of the one over four would cancel, and I would keep getting cancellation all the way until I get to eat to the one over. Okay, that will cancel. So this means this came on this one would also cancel with the term right before, however, is I go all the way to the end. I see that I have each of the one over Kaye plus one. And that won't cancel because all the other denominators are less. Think a plus one. So everything is canceled except eat of the one and then negative e to the one over. Kaye plus one saw me write that on the next page. And then we had inside the apprentices each of the one. What should just eat. However, as we take care to go to infinity this term here, that's just e. But this term over here becomes one over infinity, which is zero. And so in this case, if we go out and plug that in, we just have e minus one. And that's your final answer.