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JH
Numerade Educator

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Problem 64 Hard Difficulty

We have seen that the harmonic series is a divergent series whose terms approach 0. Show that
$ \displaystyle \sum_{n = 1}^{\infty} \ln \left( 1 + \frac {1}{n} \right) $
is another series with this property.

Answer

$\lim _{N \rightarrow \infty} \ln (N+1)-\ln 1=\infty-0=\infty$

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Video Transcript

we've seen that the harmonic Siri's, which recalled this, means the Siri's and equals one to infinity of one of her end and notice that a N equals one over and does satisfy it goes to zero as and goes to infinity. However, the Siri's the Harmonic series, The Diversions We'LL show that this given example that we have here, this son has a similar property will show that the end Well, we could actually show this right now is we let and go to infinity. This is limit and goes to Infinity Natural Log one plus one over N. And we could go out and actually take the limit. And inside of the algorithm, the reason you could do this is because the natural log is continuous and then we get natural log of just one plus zero. That's natural log of one in that equal zero. So the limit of the term and here the thing that were adding up in the summation that does go to zero. However, we'LL end up showing that in Siri's itself is divergent, and it's so in this sense that has the same properties is the harmonic series. So let's show this thing Diversions. I can rewrite this. Actually here I'Ll need more room So let me go on to the next page Here. This is our Siri's. Let me do two steps here First let me write. This is a limit as que goes to infinity And then I'll sum from one's Okay. And then also let's rewrite this is bye come getting a common denominator. So this let's re write this by using the properties of the lottery with, um, we know that Ellen of Bill Ray is Ellen B minus ln a just using that property there. So now let's go ahead and do telescope. Or, in other words, this is what the book calls telescoping Siri's. And the examples. Yeah. So this is where we've pulled out the limit. And now well, just go ahead and evaluate this term here and then at the very end, will come out and take a limit. So first, let's find the sun. Let's plug in and equals one that's our first term, then plug in and equals two and equals three and so on. And we will keep going in that direction. And then eventually, before we get okay, we would plug in K minus one. And then the very last term up here would be only plug in and equals K. So here are indicating blew. This was the end value that was plugged in to get this term to get the second term and so on. This was all the way at the very end and equals k. And this in this term right here was the term right before that. So now we look to see how much we could cancel. This is the telescoping part, Ellen of one. We know that zero. So we could ignore that. We see this positive, Ellen, to cancels with the negative positive, Ellen. Three cancers with the negative. This positive Ellen for would end up canceling with the following her, sir. And then it looks like I could keep cancelling K minus one. This would cancel with the term right before it. And then I would be able to cancel the Ellen of case. However, I'm still left over with this woman here. And in this limit, this goes to infinity. This means that the sum is diversion. So on the very love side. Look at what we started in the top left. We were trying to evaluate the entire series. We simplified it a little bit using the long properties. We telescoped, we wrote. It is a limit. Since the limit goes to infinity, That's not a real number. And since the limit is not a real number, that means that the series is divergent and that's our final answer.