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## Integral Calculus

### Course: Integral Calculus > Unit 5

Lesson 6: Comparison tests# Proof: harmonic series diverges

AP.CALC:

LIM‑7 (EU)

, LIM‑7.A (LO)

, LIM‑7.A.8 (EK)

Showing that the harmonic series 1 + ½ + ⅓ + ¼ + ... actually diverges, using the direct comparison test. This proof is famous for its clever use of algebraic manipulation!

## Want to join the conversation?

- I understand the intuition behind this proof, but I'm not fully convinced. It seems to me that as you keep adding terms to the lesser series, it takes more and more terms to add up to another 1/2. As n approaches infinity, it will take an infinite number of terms to add up to another 1/2, and the sum (it seems to me) need not necessarily keep growing. Where is the proof that that the second series diverges?(82 votes)
- Hank, your observation spurred me to find an answer myself, so I ran some simulations. Interestingly I noticed that for each increase in order of magnitude of the number of terms, the sum of the series increases by approximately 2.3, however this number seems to increase rather than decrease, indicating a non-convergence.

For example, say we take 10 terms, we end up with approximately 2.92. If we take 100 terms, we end up with 5.19. The difference between these two numbers is about 2.26.

However, if we take 100,000 terms, we end up with approximately 12.09, compare this to one order of magnitude less, which is only 10,000 terms and we end up with approximately 9.79. The increase here has gone up to 2.3.

As the order of magnitude increase approaches very large numbers (I ran a simulation going to 9 orders of magnitude, a steady increase of 2.3 is noticed. This increase by 2.3 per order of magnitude does change very minimally in a positive as opposed to a negative direction. This seems to indicate the series is diverging (if it were lowering I'd say it would eventually converge).

Was just a very simple simulation, here is the code in R:

n=desired number of terms

sum=0

for (i in 1:n) {

sum=(sum + 1/i)

}(114 votes)

- hi all! Could you please explain me what means "the largest power of 1/2"?

If it means (1/2)^x, where x is what we are looking for? Therefore,

(1/2)^0 = 1 and this is a largest power of 1/2 that is less or equal to 1.

(1/2)^1 = 1/2 and this is a largest power of 1/2 that is less or equal to 1/2

(1/2)^2 = 1/4 and this is a largest power of 1/2 that is less or equal to 1/3

Here I have a question.

So, we cannot take non-integer power hence we must take only whole numbers?

Because if we want to find the exactly large power of (1/2) that less or equal 1/3 it should be

(1/2)^ln3/ln2. This term is equal to 1/3.

Thank you, Ivan(21 votes)- Well, it's quite simple actually: If we substituted 1/3 with (1/2)^(ln3)/(ln2), both series would be the same. And the thing here is to prove any sum "lesser" than the harmonic diverges. So we could of course use rational exponents, but it wouldn't help us in any sense.(3 votes)

- Do all harmonic like series with infinite linear terms always diverge? Will something like -

1/1+1/1+1/2+1/3+1/5+1/8+1/13+1/21...

where the denominator is a Fibonacci number. Will this diverge? Proof?(10 votes)- Here is a way to think about it. The ratio between each term is the reciprocal of the golden ratio, which is less than 1, so by the ratio test, the series should converge. The great mathematician Paul Erdos conjectured the convergent sum to be irrational. This was proven by another in 1989.(16 votes)

- The series 1/n diverges, and it is really nicely shown in this video. But why for example 1/n^2 or 1/n^3 converges? If we try to find the sum, it also goes to infinity (we just have to sum more fractions than in 1/n) , so how is it possible?(8 votes)
- Hi Elsa!

First, the series 1/n and 1/² both have the same number of terms, an infinite number.

The reason that 1/n² converges has to do with how fast each term is getting smaller. That is, each term in 1/n² decreases at a sufficiently fast rate that even though you continue to add to the sum, what you are adding is so infinitesimally small that it makes no difference to the final sum.

Think about this. You are at point a. To get to point b you first need to go half the distance from a to b. So now you are at a'. Now to get to b you need to go half way from a' to b. That gets you to a''. Do you see how we can keep doing this process infinitely many times and NEVER get to b.

In a similar manner, the terms add so little to the sum that they do not impact the total. There are several proofs, one using the integral test, another uses the Cauchy Condensation test. I suggest you do a search on the convergence of 1/n² to get more background so you can feel confident in your own mind why this is so.(8 votes)

- Why do we choose powers of 1/2 in particular? Seems a little random. If we were trying to use the comparison test later on in a another situation, how would we know what fraction to use?(7 votes)
- Oresme’s proof is very witty and I believe finding such proofs requires patience, serendipity, creativity and mathematical experience.

However, I do have a recipe to approach the proof of a series’ convergence/divergence with the limit comparison test.

1) I start with an intuition: does S looks like is converges or diverges? Let’s say it looks like it converges.

2) Then I look for another series S' such that:

a) I know that S' converges. For example, a p-series with n>1, or a geometric series with |r|<1.

b) The limit of S(n)/S'(n) with n going to infinity will be finite and positive. Typically, the dominant term of the numerator should be the same as the dominant term of the denominator so that the limit of the ratio will be finite.

If I can’t find a nice S', then maybe it means that my intuition was wrong: maybe S is divergent. If so, I try again with a S' that I know is divergent (such as the harmonic series, or a geometric series with |r|>1).

For example: S = Σ(n=1 to infinity)((n+1)/(n³+6n+3)).

1) The dominant term of the denominator (n³) is larger than the dominant term of the numerator (n), so my intuition says that S should converge.

2) a) Since I am dealing with rational functions (ratios of polynomials), a p-series with n>1 would make a good S'. Those are convergent, and

b) for my Sn/S'n, I’d like to have the same dominant term in the numerator and the denominator. So if I divide Sn by 1/n², I’ll multiply it by n², and that raise the degree of the numerator to n*n², that is n³: [(n+1)/(n³+6n+3)]/[1/n²] = n²(n+1)/(n³+6n+3) = (n³+n²)/(n³+6n+3). The limit of this is 1! Yeah!

Hope this helps.(6 votes)

- Although
`1/n → 0`

as`n → ∞`

, the summation still changes if we keep adding. So the sum of a convergent series is a**fixed**value (I mean, an arbitrary number such as 171674, 2, 1/π... ) and the sum of a divergent series is not fixed, right?(6 votes)- Well, it's true for both a convergent series and a divergent series that the sum changes as we keep adding more terms. The distinction is in what happens when we attempt to find the limit as the sequence of partial sums goes to infinity. For a convergent series, the limit of the sequence of partial sums is a finite number. We say the series diverges if the limit is plus or minus infinity, or if the limit does not exist.

In this video, Sal shows that the harmonic series diverges because the sequence of partial sums goes to infinity. On the other hand we could have a geometric series that is the sum of 1+1/2+1/4+1/8+1/16+ . . .. The sum keeps changing as we add another term, but it's always getting closer and closer to a limit (which happens to be 2), so the series converges.(4 votes)

- Can't you just apply the ratio test?

1/n divided by 1/(n+1)

is (n+1)/n

n+1 is always bigger than n

so (n+1)/n is larger than 1

so it diverges!(0 votes)- You'd have to divide 1/(n+1) by 1/n and take the limit of it as n aproaches infinity, so you get 1.

The ratio test can only tell you if something diverges or converges if the limit is not equal to 1, so the ratio test is inconclusive for the harmonic series.(19 votes)

- I am really confused about this. I know the rule is, if you have an equation of 1 / (n^p)

and p is greater than 1.. then the sum converges.. but if it is equal to 1 or less.. then it diverges. I know thats the rule. But WHY?

Here is my argument. 1/(n^2) will never actually "reach" 0. Which means the sum will never technically stop increasing.

I mean our calculators say it does because they can't calculate out to that far.. But assuming we can go beyond the Planck limits, we should be able to keep going for infinity, and yes the next add on is so small, but if you graphed the infinite sum of 1/(n^2), putting the n terms on the x axis, and the total sum on the y axis, there is some n for 1000 on the y axis, and there is some n value for any y axis value right!? how is my intuition wrong here?

Or is it some sort of hazy math idea that.. "yeah.. technically you could reach any y axis value, but at some value of p it would take so long, practically it converges".. and mathematicians just defined that at any value greater than 1?

And if thats the case, WHY did the mathematicians pick values greater than 1?(4 votes)- For a proof of the convergence of any p-series where p > 1, I'd just recommend checking out the videos for the Integral Test here on Khan Academy. It'd be a bit hairy to write that here without good math notation. However, I can give you some intuition/examples as to why these sums can converge to constant values, and why they're bounded (they can't increase forever).

You're correct that the sum for a series like 1/n^2 will never stop increasing. However, it's incorrect to say that the sum will continue to increase**without bound**- there are some values that the sum simply cannot reach. Let's look at a simpler example than 1/n^2; this one isn't a p-series, but it clarifies an important concept.

Take the sum of`9 * (1/10^n)`

, starting at`n=0`

and continuing on for all integer values of n until some specified bound. Clearly, this sum is equal to`9 + 0.9 + 0.09 + 0.009 + ... = 9.999...`

. Immediately, we can see that for any finite value of n, this sum will never equal 10; it will always be`1/10^n`

less than 10. Thus, it is possible for sums to continue infinitely, and to keep growing forever, but still to be**bounded by some constant**. The terms are just consistently too small to make the total sum greater than or equal to 10.

Now look at what happens when we take the limit of the example above as n approaches infinity. The same fact holds: the final sum is`1/10^n`

less than 10. However, as n approaches infinity,`1/10^n`

approaches 0. Thus, in the limit, our sum is equal to`10 - 0`

. That is, our sum is**exactly equal to 10**in the limit. This shows that sums can "converge" to a constant value in the infinite limit.

The intuition behind the convergence of series like 1/n^2 (or in general, 1/n^p for p > 1) is the same as the intuition for the example above. Yes, the terms will keep increasing, but the overall sum is bounded - it can only increase up to a certain point.

Hope all that helps!(6 votes)

- If one were to graph 1/n they would have a curve which as it reaches infinity approaches an asymptote of zero. If it converges on zero, or any other number, does this not meet the definition of convergence? And how does any tweeking of formula ignore the fact the given formula we were initially given to evaluate converges on zero when the limit is infinity?(4 votes)
- For your consideration, Greg - you
*may*be confusing terminology - sequence vs. series.

A series is the sum of a sequence.

The limit of the**sequence**`1/n as n-->∞ is 0`

This means that as n gets bigger, each successive term in the sequence gets smaller and smaller and so, as you noted, each term gets closer and closer to 0.

BUT

The limit of the**series**`1/n as n-->∞ is infinity`

.

This means that*even though*each successive term in the sequence is getting smaller and smaller, if you keep adding them on to all of the previous terms summed so far, the sum will continue to get larger and larger with out bound since each term is not sufficiently small enough that the sum is bounded, unlike the series`1/n²`

, which is bounded and does converge.

The example Sal did was to create a new series where each term was equal to or smaller than the harmonic series and show that the new series diverges, so the harmonic series must diverge as well (this method is called the Limit Comparison Test https://www.khanacademy.org/math/calculus-home/series-calc/convergence-divergence-tests-calc/v/limit-comparison-test-cor)

Hope that helped, if not, comment and I will answer your questions as quickly as I can.(6 votes)

- How would you express the second series (S) in sigma notation?(4 votes)
- The second series is just the sum of recurring 1/2's. So it would be

Sum from n = 1 to infinity of 1/2(2 votes)

## 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.