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Problem 36 Hard Difficulty

Use the following steps to show that
$ \displaystyle \sum_{n = 1}^{\infty} \frac {( - 1)^{n-1}}{n} = \ln 2 $
Let $ h_n $ and $ s_n $ be the partial sums of the harmonic and alternating harmonic series.
(a) Show that $ s_{2n} = h_{2n} - h_n. $
(b) From Exercise $ 11.3.44 $ we have

$ h_n - \ln n \to y $ as $ n \to \infty $

and therefore

$ h_{2n} - \ln(2n) \to y $ as $ n \to \infty $

use these facts together with part (a) to show that $ s_{2n} \to \ln 2 $ as $ n \to \infty . $

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Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 11

Infinite Sequences and Series

Section 5

Alternating Series

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Let's use the following steps to show that this infinite Siri's conversions to the natural log of two. So here's some notation that's given agent That's the harmonic, some up insulin So we can write. This is I equals one upto end of one over I in s. And so these air defined earlier Some point in this chapter this's the alternating harmonic some. So we also go from one side And but then we we want this sum to start a positive one. So I'll do the and minus one and then oh oh, excuse me, I might this one. Sorry about that again. That minus one Or you could do a plus one Doesn't matter. It's just to make sure that when you plug in, I equals one that the first term is one. So, for example, this looks like one minus one over two, plus one over three, and so on. Now for part A. Here's an identity for us to verify, so let's go ahead and try to verify this by induction. So first, let he event be the proposition that has to end. Soap even is a proposition. It's It's a claim that this equation is true when end when we're using this value of n So let's do proof by induction. Mathematical induction, of course. So first we do the bass case. So this is when an equals one. Well, let's plug that in to make sure we're getting a true statement. So the question is, is as to equal to H of two minus age of one. Now as to specious one minus a half, which is a half h two is one plus a half in a church one is just one which is also a half. So yes, this is true. So that takes care of the bass case. The first half of the proof for the induction. Now we go toe what's usually the harder half. This is the inductive step. So these are the required steps that's needed when using proved by induction. So this is where we assume Pien is true And then we want to show that p of n plus one is true. So this is for any end. So by assuming Pien is true, we're making the claim that as of two n equals age of to end minus h of n now we'd like to show that if we replace and within plus one, that the equation is still true. So let's go on. I'm running out of room here. Let's go to the next page. So now we go to the inductive step, so notice that when you increase and buy one, it becomes too. And plus two. So we have a job to end plus two minus h of n plus one so we can rewrite this. So I'm going to do here is just rewrite this age of two and plus two as h of two n plus the next two terms in line in the harmonic series. So these terms in the red parentheses or equal, they're both equal to each of two and plus two. And then here, let me just write. This last part is H and plus the next term in line in the harmonic series. So that takes care of the green part and parentheses. So the things in the red parentheses are equal. The things in the blueprint to Caesar and that Excuse me, the green parentheses, air people. Now let's go ahead and just clean this up a little bit. I see that I could combine the age to end in the kitchen. So h two n minus h n By our assumption, this is just s itu in. So this is just by assumption that Pien is true So not only do I have to subtract these terms, I can replace him with us to end And then after that we still have some terms a line here we still have the plus one over two M plus one and then the minus one over and plus one and then plus one over two m plus two. However, this term down here this is just equal to us of two one plus two just by using the definition rating out s of two. And this is true because And then you would have to add in the next two terms in line. So if you go ahead and add in the next term's align while the next one would be plus one over two M plus one and then minus one over two m plus two. So we would have to come up here and show that this term right here is equal to oh, those. So here we can see that these terms match up. So the question would be when you simplify this minus one over and plus one plus one over two M plus two. Do you get this term down here? And you could just quickly add the fractions to see if that's true. So I could multiplies, happened bottom of this by two and then add the one in from this guy and you could see that's a true statement. So this justifies this equation over here. Why? I jumped down from s of two and plus two and I just went to the definition of essence to and plus two. I wrote it out. I simplified these two terms and I realize that this expression here circled in black is equal to this expression circled in red. And this verifies the induction. So we just sure p of n plus one is true. Why is that? We started with H of Tuen plus two minus H event plus one and we simplified to get s of two and plus two and we use the inductive step and this equation over here So that takes care of party that verifies the identity. Let's go on to the next page for part B. So for part B, we want to use part A and we'd like to recall a fact from this is mentioned in the problem. This is not me saying that's so. This is that h of n minus natural law given approaches some constant gamma. This has a name. This is known as oil is constant. So this is of course, as and goes to infinity. This just by replacing and with two. And this implies a chav too, and minus natural log of to end also approaches Gammas and goes to infinity. Now let's use thes fax with part, eh? To show that s of to end approaches natural log to this is what we want to show. So now let's go ahead and write this out. So s of to end Bye part, eh? It's just a job to end minus h oven. Now let me play around here and do some algebra and then plus natural log of two and minus Ellen of end. So here you could cancel out. Did you see that these terms are equal, but writing it in this way? Well, let me use the previous fax. So this is all equal to well here first not equal to I should take the limit. So let me go on to the next page here I'm gonna end up taking a limit. So you'LL want to record this side and then also the Rame Alcide And now let's go ahead and take the limit as little and goes to infinity so we can see that the first two terms in the parentheses by these facts up here we'LL both go to Gamma So that's a gamma here and then minus another gamache And then we have this term down here So let's go to the next page. So taking the limit as n goes to infinity on both sides So that's giving us Let me just go ahead. And I already said this parts I'll just replace those with my Gammas. The first two terms in the parentheses went to gamma minus gamma But then we still have the limit as n goes to infinity of natural log to end minus natural lot of end This was the very last term. Now we can just go ahead and usar los properties here Natural log of two and over too. Oh, excuse me to end over. End about a little careless there. Cancel off those ends and then you just have natural log of two. So we ended up to showing that s of Tuen goes to Ellen of too. And that terrifies R B. That's our final answer.

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Video Transcript

James Stewart

Calculus: Early Transcendentals

Anna Marie Vagnozzi

Heather Zimmers

Kayleah Tsai

Caleb Elmore

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Anna Marie Vagnozzi

Heather Zimmers

Kayleah Tsai

Caleb Elmore

James StewartCalculus: Early Transcendentals

Anna Marie Vagnozzi

Heather Zimmers

Kayleah Tsai

Caleb Elmore

James Stewart

Calculus: Early Transcendentals