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JH
Numerade Educator

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Problem 77 Hard Difficulty

In Example 9 we showed that the harmonic series is divergent. Here we outline another method, making use of the fact that $ e^x > 1 + x $ for any $ x > 0. $ (See Exercise 4.3.84.)
If $ s_n $ is the $ n $th partial sum of the harmonic series, show that $ e^{x_n} > n + 1. $ Why does this imply that the harmonic series is divergent?

Answer

$\left\{s_{n}\right\}$ is increasing, $\lim _{n \rightarrow \infty} s_{n}=\infty,$ implying that the harmonic series is drvergent.

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Video Transcript

So here we are given as an is the end partial some of the harmonic series. So that's just the sum over here. One one over one one over two, one over three, all the way to one over it. And then here we like to show e to the s. And that should be a set up there is bigger than one plus one. So let's go ahead and show this So we have even s and well, by definition, just replace sn with the sun. Oh, then using your laws of exponents, we could go ahead and rewrite. This is a product of ease. So either the one either the one half, either the one third that either the one over. Now this is where we'Ll use the previous equation from the earlier problem. The book it's gotten use this fact here and times so here since one over and is bigger than zero. We're just going to use this formula and replace all the one over eyes someone over one one over two, one over three, one over. Ed all those will be replaced with X and we'LL just use this formula here. So this is larger than one plus one one plus ass one plus a third all the way up to one plus one of red. And we could go ahead and simplifying this. So this is two over one three over too for over three. That that that and plus one over N Now here. Let's go ahead and start canceling. We could cancel the twos. We could cancel the threes. We could cancel everything all the way up until the end. But then we're left with n plus one over one. What's this? Just n plus one and this is unbounded. So this is unbounded above. So this M plus one will continue to grow. So we have just shown that E to the SN is unbounded. I shouldn't write it that way. Let me take a few steps backward. Sorry about that. We just shown the sequence. Yes, and is not abounds in above. Therefore we must have That s and so there will be our final answer after, Of course, the previous inequalities is not founded above. So basically the idea here in the bottom left is we've just shown that e to this exponents is getting very, very big. But The only way e to the X gets really, really big is if the thing and the exponents is getting really, really big. And that's what we wanted to show. We wanted to show that the exponents is getting big, and the way we did that was by showing that either the ex close to infinity and this was our previous line of reasoning in the original inequality. So this implies that the harmonic series it's not bounded above. This means that it diverges, and that's our final answer.