Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am struggling understanding intuitively why the harmonic series diverges but the p-harmonic series converges. I know there are methods and applications to prove converges, but I am only having trouble understanding intuitively why it is. I know I must never trust my intuition, but this is hard for me to grasp. In both cases, the terms of the series are getting smaller, hence are approaching zero, but they both result in different answers. $$\sum_{n=1}^{\infty}\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots = \text{diverges}$$ $$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots =\text{converges}$$

share|improve this question
4  
To make matters worse, $$\sum_{n=1}^\infty\frac{1}{n^{1+\epsilon}}$$ converges for all $\epsilon > 0$. That might make it a bit hard to "intuitively" grasp. –  EuYu Apr 20 '13 at 7:25
    
“Young man, in mathematics you don't understand things. You just get used to them.” ― John von Neumann –  Leo Apr 20 '13 at 8:49
    
I came looking for an answer trying to cope with this: en.wikipedia.org/wiki/Ant_on_a_rubber_rope –  yati sagade Nov 5 '13 at 7:10

6 Answers 6

up vote 11 down vote accepted

Firstly, you should always use your intuition. If you find that your intuition was correct, then smile. If you find that your intuition was wrong, use the experience to fine-tune your intuition.

I hope I'm interpreting you question correctly - here goes. Since you are not interested in any of the proofs, I'll just focus on intuition. Now, let's consider a series of the from $\sum _n \frac{1}{n^p}$, with $p>0$ a parameter. Intuitively, the convergence or divergence of the series depends on how fast the general term $\frac{1}{n^p}$ tends to $0$. This is so because the sum is that of infinitely many positive quantities. If these quantities converge to $0$ too slow, the number of summands in each partial sum will be more dominant than the magnitude of the summands. However, if the quantities converge to $0$ fast enough, then in each partial sum the magnitude of the summands will be dominated by numbers of small magnitude, and thus outweigh the fact that there are lots of summands.

So, the question is how fast does $\frac{1}{n^p}$ converge to $0$. Let's look at some extreme values of $p$. If $p$ is very large, say $p=1000$, then $\frac{1}{n^p}$ becomes very small very fast (experiment with computing just a few values to see that). So, when $p$ is large, it seems the general term converges to $0$ very fast, and thus we'd expect the series to converge. However, if the value of $p$ is very small, say $p=\frac{1}{1000}$, then $\frac{1}{n^p}$ is actually pretty large for the first few possibilities for $n$, and while it does monotonically tend to $0$, it does so very slowly. So, we'd expect the series to diverge when $p$ is small.

Now, if $0<p<q$ then $\frac{1}{n^q}<\frac{1}{n^p}$, so the bigger the parameter the faster the convergence of the general term to $0$ gets. So, small values for the parameter imply diverge of the series, while large values of the parameter imply convergence of the series. So, somewhere in the middle there has to be a value $b$ for the parameter such that if $p<b$ then the series diverges, while if $p>b$ then the series converges.

So, just by this straightforward analysis of the behaviour with respect to varying the parameter $p$, we know (intuitively) that there must be some cut-off value for $p$ that is the gateway between convergence and divergence. What happens at the that gateway value for $p$ is unclear, and there is no compelling reason to suspect one behaviour of the series over another. Now, the particular whereabouts of that special gateway value for $p$ should depend strongly on the particularities of the general term. This is thus where you'll have to delve into more rigorous proofs.

I hope this rather lengthy answer addresses what you were wondering about. Basically, it says that a cutoff parameter must exist, but we can't expect to say anything about its whereabouts nor the behaviour at that cutoff value without careful study of the general terms.

share|improve this answer
    
Thank you very much! –  Lays Apr 21 '13 at 3:21
    
you are very welcome :) –  Ittay Weiss Apr 21 '13 at 3:27

Intuitively the main argument why the harmonic series diverge is that $\forall k \sum_{n=k}^{n=2k}\frac{1}{n}>k\frac{1}{2k}=\frac{1}{2}$ since smallest element is $\frac{1}{2k}$ and there are k elements in the interval $[k;2k]$. So the harmonic sum for any finite interval $[k;2k]$ is > 0.5.
So if you split the infinite interval over which you make the summation into intervals $[1;k][k;2k][2k;4k],...,[2^nk;2^{n+1}k],...$ each interval has sum higher then 0.5 and since there are infinite number of this intervals to cover the whole $\mathbb{Z}$ it diverges.
Basically they get smaller and smaller, but not fast enough to converge to a limit. The p-harmonic on the other hand because of the square in the denominator can not have this "ability" and converge, aka they get smaller faster enough. To try to explain it on intuitive level better thing like this: you are adding infinite number of the sequence, so in order to converge to a limit L they have to get smaller with some "speed". Now even if they get close to 0 if the speed at which they grow smaller is not high enough then they still will be too much stuff added and will never converge.

share|improve this answer
    
I dont understand your first paragraph but your second paragraph is good. When you said it is not getting small fast enough that made a little sense to me. I still have issues though understanding, maybe I shouldn't think far past it then what it actually is and just accept that fact it is divergent. Thanks! –  Lays Apr 20 '13 at 7:35
    
Tried to make the first one a bit more clear. It is basically demonstrating why the harmonics dont get smaller fast enough. –  Belov Apr 20 '13 at 7:40
    
Thank you for that! –  Lays Apr 20 '13 at 7:42
    
That inequality opened up my brain :) –  yati sagade Nov 5 '13 at 7:09

We produce two series that are close in spirit to the series you mentioned. Perhaps the divergence of the first, and the convergence of the second, will be clearer.

Consider the series $$\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\cdots.$$ So there is $1$ term equal to $\frac{1}{2}$, then a block of $2$ each equal to $\frac{1}{4}$, then a block of $4$ each equal to $\frac{1}{8}$, then a block of $8$ each equal to $\frac{1}{16}$, and so on forever. Each block has sum $\frac{1}{2}$, so if you add enough terms, your sum will be very big. But it will take an awful lot of terms to add up to $1000$, many more more terms than there are atoms in the universe. Note that each term is less than the corresponding term in the harmonic series, so if you add together enough terms of the harmonic series, the sum will be very big.

Now consider the series $$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{4^2}+\frac{1}{4^2}+\frac{1}{4^2}+\frac{1}{8^2}+\frac{1}{8^2}+\frac{1}{8^2}+\frac{1}{8^2}+\frac{1}{8^2}+\cdots.$$ Each term is $\ge$ the corresponding term in the series $1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\cdots$.

Again, we find the sums of the blocks. The first block has sum $1$. The second has sum $\frac{1}{2}$. The third has sum $\frac{1}{4}$. The fourth has sum $\frac{1}{8}$, and so on forever. So if we add up "all" the terms, we get sum $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$, a familiar series with sum $2$.

share|improve this answer
    
Thanks for this! –  Lays Apr 21 '13 at 3:20

If you convert the sum to an integral, $$ \int_1^\infty 1/x^2 dx = 1/x|_1^\infty = 1/\infty - 1 $$ converges, but $$ \int_1^\infty 1/x dx = ln x|_1^\infty = ln \infty $$ doesn't.

share|improve this answer
    
Yes, by the integral test it shows why it does converge and diverge but that isn't what I am having trouble understanding. I am having trouble understanding intuitively why it is. –  Lays Apr 20 '13 at 7:27

Hmm, I like the answer of @Andre Nicolas much, so I would like to also generalize that idea to arbitrary real exponents p in the denominator.

First let's state, that a lower bound for any such series with real exponents p is 1 so let's write $L_p=1$.

Second let's restate the upper bounding series $U_2$ for exponent $p=2$ where the repeated terms are written as multiplications as given by Andre $$ U_2 = 1 + 2[1/2^2] + 4[1/4^2] + 8[1/8^2] +... $$

Now let's write this explicitely as powers of base 2: $$ U_2 = 1 + 2^1[1/2^2] + 2^2[1/2^{2 \cdot 2}] + 2^3[1/2^{3 \cdot 2}] +... $$ We see, how this generalizes to some exponent p: $$ U_p = 1 + 2^1[1/2^p] + 2^2[1/2^{2 \cdot p}] + 2^3[1/2^{3 \cdot p}] +... $$ Now let's collect/cancel the exponents $$ U_p = 1 + 2^{1-p} + 2^{2(1-p)} + 2^{3(1-p)} +... $$ But this is now a geometric series $$ U_p = { 1\over 1-2^{1-p} } $$ and this has a finite value for any $p=1+\epsilon$ and is infinite for $\epsilon=0$ or $p=1$ And if the upper bounding series has a positive finite value, and the lower bound is 1 then the series in question must be convergent with a value in between.

share|improve this answer
    
Thank you for this! –  Lays Apr 21 '13 at 3:20

Pick any i > 0 and pick a number 0 < a < 1. You can always find an n, such that: $$ \sum_{k=i}^n \frac{1}{k} > \sum_{k=i}^n \frac{1}{n} = \frac{n-i}{n} = a $$ because the limit of the latter goes to 1. This means that you can pick any term to start and you can always go far enough to the right to find it increase by almost 1. This means the series must diverge because you can just add another 1.

The same argument doesn't hold for the p-harmonic series because: $$ \frac{n-i}{n^2} \rightarrow 0~(for~n>i \rightarrow \infty) $$

Although this is no proof it gives the intuition: the harmonic series decreases as fast (factor n) as it adds new terms (factor n). The p-harmonic series decreases faster...

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.