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Can someone give a simple explanation as to why the harmonic series

$\sum_{n=1}^\infty\frac1n=\frac 1 1 + \frac 12 + \frac 13 + \cdots $

doesn't converge, on the other hand it grows very slowly?

I'd prefer an easily comprehensible explanation rather than a rigorous proof regularly found in undergraduate textbooks.

  • 0
    Here is the funny write up for what Oiler mentioned, given by terry tao: http://mathoverflow.net/q/447422016-10-14

25 Answers 25

146

Let's group the terms as follows:

Group $1$ : $\displaystyle\frac11\qquad$ ($1$ term)

Group $2$ : $\displaystyle\frac12+\frac13\qquad$($2$ terms)

Group $3$ : $\displaystyle\frac14+\frac15+\frac16+\frac17\qquad$($4$ terms)

Group $4$ : $\displaystyle\frac18+\frac19+\cdots+\frac1{15}\qquad$ ($8$ terms)

$\quad\vdots$

In general, group $n$ contains $2^{n-1}$ terms. But also, notice that the smallest element in group $n$ is larger than $\dfrac1{2^n}$. For example all elements in group $2$ are larger than $\dfrac1{2^2}$. So the sum of the terms in each group is larger than $2^{n-1} \cdot \dfrac1{2^n} = \dfrac1{2}$. Since there are infinitely many groups, and the sum in each group is larger than $\dfrac1{2}$, it follows that the total sum is infinite.

This proof is often attributed to Nicole Oresme.

  • 0
    "This proof is often attributed to Nicole Oresme." $\leftarrow$ I attribute it to The Number Devil.2013-03-15
43

There is a fantastic collection of $20$ different proofs that this series diverges. I recommend you read it (it can be found here). I especially like proof $14$, which appeals to triangular numbers for a sort of cameo role.


EDIT

It seems the original link is broken, due to the author moving to his own site. So I followed up and found the new link. In addition, the author has an extended addendum, bringing the total number of proofs to 42+.

  • 0
    Apparently, the list has been [updated](http://prairiestate.edu/skifowit/harm2.pdf).2013-06-19
27

The answer given by AgCl is a classic one. And possibly pedagogically best; I don't know.

I also like the following argument. I'm not sure what students who are new to the topic will think about it.

Suppose 1 + 1/2 + 1/3 + 1/4 + ... adds up to some finite total S. Now group terms in the following way:

$1 + \frac{1}{2} > \frac{1}{2} + \frac{1}{2} = \frac{2}{2} = 1$

$\frac{1}{3} + \frac{1}{4} > \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$

$\frac{1}{5} + \frac{1}{6} > \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$

Continuing in this way, we get $S > S$, a contradiction.

26

This was bumped, so I'll add a proof sweet proof I saw in this site. Exponentiate $H_n$ and get $e^{H_n}=\prod_{k=1}^n e^{1/k}\gt\prod_{k=1}^n\left(1+\frac{1}{k}\right)=n+1.$ Therefore, $H_n\gt\log(n+1)$, so we are done. We used $e^x\gt1+x$ and telescoped the resulting product.

  • 1
    Oh, that's unique.2016-10-07
24

Let's group the terms as follows:$A=\frac11+\frac12+\frac13+\frac14+\cdots\\ $ $ A=\underbrace{(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{9})}_{\color{red} {9- terms}} +\underbrace{(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\cdots+\frac{1}{99})}_{\color{red} {90- terms}}\\+\underbrace{(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+\cdots+\frac{1}{999})}_{\color{red} {900- terms}}+\cdots \\ \to $ $\\A>9 \times(\frac{1}{10})+(99-10+1)\times \frac{1}{100}+(999-100+1)\times \frac{1}{1000}+... \\A>\frac{9}{10}+\frac{90}{100}+\frac{90}{100}+\frac{900}{1000}+...\\ \to A>\underbrace{\frac{9}{10}+\frac{9}{10}+\frac{9}{10}+\frac{9}{10}+\frac{9}{10}+\frac{9}{10}+...}_{\color{red} {\text{ m group} ,\text{ and} \space m\to \infty}} \to \infty $

Showing that $A$ diverges by grouping numbers.

22

An alternative proof (translated and adapted from this comment by Filipe Oliveira, in Portuguese, posted also here). Let $ f(x)=\ln(1+x)$. Then f'(x)=\dfrac {1}{1+x} and f'(0)=1. Hence

$\displaystyle\lim_{x\to 0}\dfrac{\ln(1+x)}{x}=\lim_{x\to 0}\dfrac{\ln(1+x)-\ln(1)}{x-0}=1,$

and

$ \displaystyle\lim_{n\to\infty} \dfrac{\ln\left(1+\dfrac{1}{n}\right)}{\dfrac {1}{n}}=1>0.$

So, the series $\displaystyle\sum\dfrac{1}{n}$ and $\displaystyle\sum\ln\left(1+\dfrac {1}{n}\right)$ are both convergent or divergent. Since

$\ln\left(1+\dfrac {1}{n}\right)=\ln\left(\dfrac{n+1}{n}\right)=\ln (n+1)-\ln(n),$

we have

$\displaystyle\sum_{n=1}^N\ln\left(1+\dfrac {1}{n}\right)=\ln(N+1)-\ln(1)=\ln(N+1).$

Thus $\displaystyle\sum_{n=1}^{\infty}\ln\left(1+\dfrac {1}{n}\right)$ is divergent and so is $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n}$.

20

This is not as good an answer as AgCl's, nonetheless people may find it interesting.

If you're used to calculus then you might notice that the sum $ 1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}$ is very close to the integral from $1$ to $n$ of $\frac{1}{x}$. This definite integral is ln(n), so you should expect $1+\frac{1}{2}+\frac{1}{3}+\dots+ \frac{1}{n}$ to grow like $\ln(n)$.

Although this argument can be made rigorous, it's still unsatisfying because it depends on the fact that the derivative of $\ln(x)$ is $\frac{1}{x}$, which is probably harder than the original question. Nonetheless it does illustrate a good general heuristic for quickly determining how sums behave if you already know calculus.

  • 0
    The sum is closer to the integral from $\frac{1}{2}$ to $n+\frac{1}{2}$ of $\frac{1}{x}$, which is $log(2n+1)$ http://math.stackexchange.com/a/1602945/1347912016-01-25
16

Another interesting proof is based upon one of the consequences of the Lagrange's theorem applied on $\ln(x)$ function, namely:

$\frac{1}{k+1} < \ln(k+1)-\ln(k)<\frac{1}{k} \space , \space k\in\mathbb{N} ,\space k>0$

Taking $k=1,2,...,n$ values to the inequality and then summing all relations, we get the required result.

The proof is complete.

13

There also exists a proof for the divergence of the harmonic series that involves the Integral Test. It goes as follows.

It is possible to prove that the harmonic series diverges by comparing its sum with an improper integral. Specifically, consider the arrangement of rectangles shown in the figure of $ y = \dfrac {1}{x} $:

y=1/x

Each rectangle is $1$ unit wide and $\frac{1}{n}$ units high, so the total area of the rectangles is the sum of the harmonic series: $ \displaystyle\sum \left( \text {enclosed rectangle are} \right) = \displaystyle\sum_{k=1}^{\infty} \dfrac {1}{k}. $Now, the total area under the curve is given by $ \displaystyle\int_{1}^{\infty} \dfrac {\mathrm{d}x}{x} = \infty. $Since this area is entirely contained within the rectangles, the total area of the rectangles must be infinite as well. More precisely, this proves that $ \displaystyle\sum_{n=1}^{k} \dfrac {1}{n} > \displaystyle\int_{1}^{k+1} \dfrac {\mathrm{d}x}{x} = \ln (k+1). $This is the backbone of what we know today as the integral test.

Interestingly, the alternating harmonic series does converge: $ \displaystyle\sum_{n=1}^{\infty} \dfrac {(-1)^n}{n} = \ln 2. $And so does the $p$-harmonic series with $p>1$.

10

$\int_{0}^{\infty}e^{-nx}dx=\frac1n$

$\sum_{n=1}^{\infty}\int_{0}^{\infty}e^{-nx}dx=\lim_{ m \to \infty}\sum_{n=1}^{m}\frac1n$

using the law of Geometric series

$\int_{0}^{\infty}(\frac{1}{1-e^{-x}}-1)dx=\lim_{ m \to \infty}H_m$

$\lim_{ m \to \infty}H_m=\left [ \ln(e^x-1)-x \right ]_0^{\infty}\to\infty$

  • 0
    Oh right, duh, didn't quite use that FTOC correctly.2016-10-07
10

Suppose to the contrary that converges.

Let $s_n$ denote the $n$-th partial sum. Since the serie converges, $(s_n)$ is a Cauchy sequence. Let $\varepsilon = 1/3$, then there is some $n_0$ such that $|s_q-s_p|< 1/3$ for all $q>p\ge n_0$. Let $q=2n_0$ and $p=n_0$. Then

$\frac{1}{3}>\bigg|\sum_{n=n_0+1}^{2n_0} \frac{1}{n}\bigg|\ge\bigg|\sum_{n=n_0+1}^{2n_0} \frac{1}{2n_0}\bigg|=\frac{1}{2}$

a contradiction. Then this contradiction shows that the series diverges.

9

Another (different) answer, by the Cauchy Condensation Test :

$\sum_{n=1}^\infty \frac{1}{n} < \infty \iff \sum_{n=1}^\infty 2^n \frac{1}{2^n} = \sum_{n=1}^\infty 1< \infty $

The latter is obviously divergent, therefore the former diverges. This is THE shortest proof there is.

9

Let's assume that $\sum_{n=1}^{\infty}\frac1n=:H\in \mathbb{R}$, then $H=\frac11+\frac12+\frac13+\frac14+\frac15+\frac16 +\ldots $ $H\geqslant \frac11+\frac12 +\frac14+\frac14+\frac16+\frac16+\ldots$ $H\geqslant \frac11+\frac12+\frac12+\frac13+\frac14+\frac15+\ldots$ $H\geqslant \frac12 +H \Rightarrow 0\geqslant \frac12$ Since the last inequality doesn't hold, we can conclude that the sum doesn't converge.

8

enter image description here

First suppose $\displaystyle A=\frac11+\frac12+\frac13+\frac14+\cdots$ converges then show that $A>A$. That's paradox.

  • 2
    I am afraid that this approach is incorrect, since similar versions of it can be applied to *convergent* infinite series. The reason that it does not work is because the number of terms is infinite. All you've proven is that the second series approaches the final value *faster* than the first one. But whether this value is ultimately finite or not, you have not shown.2015-01-14
8

A non-rigorous explanation I thought of once: consider a savings scheme where you put a dollar in your piggy bank every day. So after $n$ days, you have $n$ dollars; clearly, your savings approach infinity. On the other hand, each day you add an additional $1/n$ proportion of your existing savings, "so" (the non-rigorous step) the accumulated percentage after $n$ days is $1 + 1/2 + \cdots + 1/n$.

This can be made rigorous through the infinite product argument $\prod_{n = 1}^\infty (1 + \tfrac{1}{n}) < \infty \iff \sum_{n = 1}^\infty \frac{1}{n} < \infty$ which is obtained, essentially, by taking the logarithm of the left-hand side and using the power series for $\log (1 + x)$.

7

Another answer that's very similar to others. But it's prettier, and perhaps easier to understand for the 9-th grade student who asked the same question here.

The student's question was ... does the sum equal some number $S$. But, look:

enter image description here

So, whatever it is, $S$ is larger than the sum of the infinite string of $\tfrac12$'s shown in the last line. No number can be this large, so $S$ can't be equal to any number. Mathematicians say that the series "diverges to infinity".

6

I think the integral test gives the most intuitive explanation. Observe that $\int^n_1 \frac1x dx= \log n$ The sum $\displaystyle\sum^n_{k=1}\frac1k$ can be viewed as the area of $n$ rectangles of height $\frac1k$, width $1$ (with the first one having it's left hand side on the y axis, and all having their bottom on the x axis). The graph of $x\mapsto \frac1x$ can be drawn under these, so the sum will grow with $n$ at (least) as fast as the integral - hence will grow (at least) logarithmically.

  • 0
    use$\log$to get nice formatting for $\log$2013-11-01
5

Let be the partial sum $H_n = \frac11 + \frac12 + \frac13 + \cdots + \frac1n$. Using Cesàro-Stolz: $ \lim_{n\to\infty}\frac{H_n}{\log n} = \lim_{n\to\infty}\frac{H_{n+1}-H_n}{\log(n+1)-\log n} = \lim_{n\to\infty}\frac{\frac1{n+1}}{\log(1+1/n)} = \lim_{n\to\infty}\frac{\frac1{n+1}}{\frac1n} = 1 $ and $\sum_{n=1}^\infty\frac1n = \lim_{n\to\infty}H_n = \infty.$

3

We all know that $\sum_{n=1}^\infty\frac1n=\frac 1 1 + \frac 12 + \frac 13 + \cdots $ diverges and grows very slowly!! I have seen many proofs of the result but recently found the following: $S =\frac 1 1 + \frac 12 + \frac 13 +\frac 14+ \frac 15+ \frac 16+ \cdots$ $> \frac 12+\frac 12+ \frac 14+ \frac 14+ \frac 16+ \frac 16+ \cdots =\frac 1 1 + \frac 12 + \frac 13 +\cdots = S.$ In this way we see that $S > S$.

  • 0
    I have saved this to my personal _The Book_ :) That being said... Come on! The last inequality itself is proof enough!! :P2016-12-12
2

Using Euler's form of the Harmonic numbers,

$\sum_{k=1}^n\frac1k=\int_0^1\frac{1-x^n}{1-x}dx$

$\begin{align} \lim_{n\to\infty}\sum_{k=1}^n\frac1k & =\lim_{n\to\infty}\int_0^1\frac{1-x^n}{1-x}dx \\ & =\int_0^1\frac1{1-x}dx \\ & =\left.\lim_{p\to1^+}-\ln(1-x)\right]_0^p \\ & \to+\infty \end{align}$


Using the Taylor expansion of $\ln(1-x)$,

$-\ln(1-x)=x+\frac{x^2}2+\frac{x^3}3+\frac{x^4}4+\dots$

$-\ln(1-1)=1+\frac12+\frac13+\frac14+\dots\quad\ $


Using Euler's relationship between the Riemann zeta function and the Dirichlet eta function,

$\begin{align} \sum_{k=1}^\infty\frac1{k^s} & =\frac1{1-2^{1-s}}\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k^s} \\ \sum_{k=1}^\infty\frac1k & =\frac10\sum_{k=1}^\infty\frac{(-1)^{k+1}}k\tag{$s=1$} \\ & \to+\infty \end{align}$

  • 0
    Yes, but since the limit as $x\to1$ in $x^n$ is monotone, it equals the asked series, if they exist.2018-03-19