Main content

## Comparison tests

Current time:0:00Total duration:6:33

# Proof: harmonic series diverges

AP.CALC:

LIM‑7 (EU)

, LIM‑7.A (LO)

, LIM‑7.A.8 (EK)

## Video transcript

- [Voiceover] This is a
painting of Nicole Oresme. I looked up the phonetic spelling
before making this video. I'm assuming I'm still mispronouncing it. But I apologize to all the French speakers out there ahead of time. He was a famous French
philosopher mathematician, who lived in medieval France,
he lived in the 1300s. And he is famous for his proof that the harmonic series
actually diverges. And just as a little bit of
review, this is a harmonic series. One plus 1/2, plus 1/3,
plus 1/4, plus 1/5. And it's always been in my brain, the first time that I
saw the harmonic series, it wasn't obvious to me whether
it converged or diverged. It looks like, well, gee,
all these terms are positive, but they're going towards
zero, so I could imagine that this thing could converge. But he proved it otherwise. He proved one of the most
famous and most elegant proofs in mathematics that it
does indeed diverge. And the way that he did this, is he replaced every term
in the harmonic series with a term that is less
than or equal to that term. And then by proving that
his new series diverges and it's less than or
equal to this series, or each of the terms
are less than or equal to each of the
corresponding terms up here, then he says, therefore,
by the comparison test, this must diverge. So, how did he construct that? One way to think about it is
he replaced each of the terms in the harmonic series with
the largest power of 1/2 that is less than or equal to that term. So, what's the largest power of 1/2 that is less than or equal to one? Well, one is a power of 1/2, so that is, 1/2 to the zero power is one. So, one is the largest power of 1/2 that is less than or equal to one. I'll just write the one there. And now, what's the largest power of 1/2 that is less than or equal to 1/2? That's just going to be 1/2, that's just 1/2 to the first power. Now, what's the largest power of 1/2 that is less than or equal to 1/3? 1/2 is larger than 1/3,
it's not less than 1/3. We want it to be less than 1/3,
so the largest power of 1/2 that is less than or equal to 1/3 is 1/4. I replace 1/3 with 1/4 and, of
course, replace 1/4 with 1/4. And then we get to 1/5. What's the largest power of 1/2 that is less than or equal to 1/5? Once again, 1/4 is greater than 1/5. The largest power of 1/2 that is less than or equal to 1/5 is 1/8. So, you replace that with 1/8. Of course, we replace for the
same reason that with 1/8. You would replace that one with 1/8. And, of course, 1/8. The largest power of 1/2 that is less than or equal to 1/8 is 1/8. And then what would you replace 1/9 with? You would replace 1/9 with 1/16
by the exact same argument. And you would keep going all
the way until you get to 1/16, so you would extensionally
have eight 1/16s in a row. What's interesting here? Let's first verify that we can use the comparison test here. In this first series each of
the terms are non-negative. In the second series each of
the terms are non-negative. And we also see that each
of the corresponding terms in a harmonic series is greater than or equal to the corresponding
term in this series. We constructed it this way. These are equal, this one is equal, this is greater than this,
1/3 is greater than 1/4. 1/4 is equal to 1/4,
1/5 is greater than 1/8, 1/6 is greater than 1/8,
1/7 is greater than 1/8. 1/8 is equal to 1/8. One way to think about it is, each of the corresponding terms in this new constructed
series are smaller. Let me just call it S. In this infinite sum, and, of course, we keep going on and on and on. Maybe i should do that in magenta. We see that each of the
corresponding terms here are smaller than the
corresponding terms up here. And they are all positive. If we can prove that S, this
sum right over here, diverges, then, by the comparison
test, the larger series, the harmonic series here, the one where the
corresponding terms are larger, that must also diverge. And how do we do that? Let's just actually take these sums. This is going to be, let me write it. S is going to be equal to one plus 1/2. 1/4 plus 1/4, what's that? That's 2/4, or 1/2. Did you see what's going on here? This is exciting. What's 1/8 plus 1/8 plus 1/8 plus 1/8? That's 4/8, or 1/2. What's 1/16 plus 1/16, and we
are going to go all the way until we are going to have eight of these. That's going to be 8/16, or 1/2. And then you are going to
have sixteen 1/32nds, or 1/2. And so, we are essentially
just going to be adding 1/2. We start with one, we
just keep adding plus 1/2, plus 1/2, plus 1/2, plus 1/2. This is clearly going to be
equal to, or this is unbounded. You could say, this is equal to infinity. Or, another way to think about
this, is S clearly diverges. And since each of its terms are smaller than the corresponding terms
in the harmonic series, we can then say the
harmonic series diverges. There is no way that this
thing over here can converge. If each of its corresponding
terms are smaller, you could even think of
the sum as being smaller. But this sum goes to infinity, so this one must also go to infinity. Hopefully, you found that
as interesting as I did.