In this video, I show that the harmonic series, 1 + 1/2 + 1/3 + 1/4 + ..., diverges by showing its even power partial sums are greater than a similar but smaller series involving repeating 1/2 term...
Transcript
Example 7 states show that the harmonic series is divergent, and the harmonic series is over here. The summation from n equals 1 to infinity of 1 divided by n equals to 1 plus 1 over 2, plus 1 over 2, plus 1 over 3 plus 1, plus, et cetera. So you keep adding another term. Yeah, so this is not a geometric series, because there's no common ratio that you're multiplying by. Instead, you're just increasing the bottom incrementally. So for this particular series, it's convenient to consider the partial sums S2, S4, S16, S32, so these powers of two, and show that they become large, and I'll show you why this is actually a very ingenious solution from my calculus book. So what will write is, I'll write here, S1, let's see right over here, so S1. equals to 1. So we have S1, initially, is the first partial sum, so where N is equal to 1, and we just have 1 over 1, is just 1, like that. So now the next term is S2, this equals 2, yeah, so once we got the first one, I'll write down S2, S4, so S2, this equals to, yeah, S2, this equals to, yeah, S1 plus the second term, yeah, so it's going to be 2 in there, so 1 over 2, 1 over 2 like that, and now S4, So what I'll do here is also write S, this is S1 to power of, or S2 to power of 0, equals to S1. And this one right here is S, just to show you that its powers, S2 to the power of 1, like that. This one here is S2 to power of 2 equals S4, like that. And going further. So this equals 2, well, we add the, you know, we just add the previous terms. So we're going to write here 1. plus one half, plus now we need to add the 1-3 and then 1-4, so 1-3 plus 1-4, put this in bracket. Okay, just put a bracket like that. And the reason I'm doing this way is because if we were to replace this 3 with a 4, what we end up having is, well, this value on the left is going to be greater than 1 plus 1 .5 plus 1 over 4 plus 1 over 4. Because here it is 1 over 3 is greater than 1 over 4. So that's the only thing we change, so then this has to be greater than this way. And this is a pretty ingenious solution method. So this, what we have here, this one over 2. So this right here is just 1 half. So in other words, 1 plus 1 half, 1 plus 1 half, and then there's a 1 plus 1 half, that's just 1 or 2 over 2. Let's put it like that for general. And now go skip over to S8, so S2 to the power of 3 equals to S8 equals 2. So if you write this whole term out, it's going to be 1 plus, again, 1 1 1.5, plus, put this in bracket, 1 over 3, plus 1 over 4, and then plus, put this in a second bracket, indicating we're going to do the same thing, but for a different. terms. So we're going to have 1 over 5 plus 1 over 6 plus 1 over 7 because we're just adding all the previous terms. So S8's going to be the summation of all of them up to S8, up to 1 over 8 like that. And here, if you can see the pattern, we're going to replace all this 5, 6, 7 with 8, so that this value right here is greater than, and we're going to write the exact same thing. 1 plus, 1 1 1⁄2, plus, 1 over 3, actually we do the same thing for this one as well. So 1 over 4, plus 1 over 4, plus here, so yeah, because we're dividing by even more than that, so these are all going to be greater than these terms, 1 over 8, plus 1 over 8, plus 1 over 8. So in other words, now this is, yeah, this is again, that's 1 half. This is 1� over here, so we have 4 over 8. This is 4 over 8, goes one-half as well. So they're all just one-halfs. So that's what we have is this equals to 1 plus 1-half plus 1-half there. So this all equals 2. Well, there's 3-1s there, so 1 plus 3 over 2. And that's why we have this 2 over 2. This is 3 over 2 now. And you can see the pattern. We're going to do the same thing now for S to the power of 4 to get S16. So we have S4 or S2, to power 4 equals to S16. So this equals to, and this equals to 1 plus 1 half, plus 1 over 3, put a bracket, 1 over 4, and then plus over here going further. 1 over 5 and also put a dot dot dot to save time and space so we have 1 over 8 like that plus we do exactly the same thing 1 over 9 plus dot dot dot all the way to 1 over 16 so that the number that we're adding from here to there is going to be 8 so it's half like that so this is going to be well greater than greater than 1 plus 1 half plus 1 over 4, plus 1 over 4, plus, then here's going to be 1 over 8, plus dot, dot, plus 1 over 6. Plus, now here is going to be 1 over 16. And we're going to have 8 of those, so we have 1 over 16, like that. All right, so now this equals 2, well, 1 plus 1.5, plus this is going to be 1⁄2. plus one half here, and then plus here is going to be 8 over 16 over 1 half, like that. So in other words, this equals 2, well, 1, and then there should be 4, 1 1⁄2, so 4 halves, so we have 1 plus 4 over 2, so we got 1 plus 2 2 2 plus 2 over 2, 1 plus 4 over 2. Yeah, so thus for the next term, the S32, so thus for S2 to the power of 5, which is going to be equals to S32, this equals to, now we're just going to have 1 plus 5 over 2. And likewise, and likewise for S2 to 6, which equals to S 64. Yeah, so the 6 is Eagle S64. And actually, yeah, this one here is greater than, am I bet, not equal to. Because notice, the sum is greater than when you simplified by adding, making these halves, so that they're greater. So, in other words, this one's going to be greater than 1 plus n noun over 2. I mean, you have 1 plus 6 over 2. I'll get to that. Because you notice here, there's a 5, there's a 5 there, there's a 6, there's a 6, there. So yes, this shows that S2N, or actually before I get to that, I'll just put it over here, so now you put it dot, dot, dot, all the way to S2N is equal to S2. Actually, I wouldn't have to read equal to, just leave it like that. S2N is greater than 1 plus n over 2. So it's going to be greater than that. So in other words, this one's going to be diverging. So circle this like that, and yeah, going back here. So this shows that S2N approaches infinity. This approaches infinity as n approaches infinity. If this is approaching infinity, we're going to have 1 plus infinity over 2, in other words, infinity, but this is greater than infinity, so it's also infinity. So thus, yeah, thus this is S2 to power of n going to infinity, but that just means that's just the even power terms. But thus in general, just s n itself is going to infinity, because you can have s n greater than s2n, and yeah, keeps going on and on, so thus this is divergent. Yeah, so thus s n, the partial sum is divergent, and therefore... or the partial sums sequence of the sums is divergent and therefore the harmonic series diverges. Note that this method of showing that the harmonic series diverges, according to my calculus book, is due to the French scholar Nicolorsmi from 1323 to 1382. Interesting, side note there.