# Why does the series $\sum_{n=1}^\infty\frac1n$ not converge?

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.

• This is not meant to be an answer but an interesting note. Suppose we denote $H(n) = 1/1 + 1/2 + ... + 1/n$ then $H(n!) - H((n-1)!) \approx log(n)$ for large n. Does this give a hint? ;) – Roupam Ghosh Jul 11 '11 at 4:14
• Here is a weakly related question: What is a textbook, or even a popularization for the general public, that (1) discusses infinite series, but (2) does not have an explanation for the divergence of this exact series? – GEdgar Nov 3 '13 at 19:50
• to avoid defining the logarithm, use the Cauchy condensation test to show that $\sum 1/n$ converges iff $\sum 1$ converges – reuns Jan 30 '16 at 23:31
• These are two of my favourite papers: The Harmonic Series Diverges Again and Again and More Proofs of Divergence of the Harmonic Series. See these. – user249332 Mar 9 '16 at 22:56
• If it converges, then it contradicts the dominated convergence theorem. This proof is easily comprehensible if you know the dominated convergence theorem, but that theorem is not the most comprehensible. – Oiler Oct 6 '16 at 23:08

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.

• +1: This is nice: it's easy to turn this into a rigorous proof, and it even gives you a lower bound for the order of growth! – Simon Nickerson Jul 21 '10 at 5:19
• I assume you mean that group 4 as 8 terms? Or do you mean to go all the way to 1/23? – Tomas Aschan Jul 21 '10 at 7:37
• Is there a closed-form function for this value? – John Gietzen Jul 21 '10 at 18:29
• Interestingly, this proof goes as far back as Nicole Oresme in the 14th century. Wikipedia has a nice display of this proof [en.wikipedia.org/wiki/Harmonic_series_%28mathematics%29] – Neil Mayhew Jul 22 '10 at 13:20
• @John: There's no explicit closed-form, but they're generally known as the Harmonic Numbers; there are a number of identities involving them (how to sum them or sum multiples of them, etc.) – Steven Stadnicki Jul 10 '11 at 21:23

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+.

• Proof 6 is also nice. – leonbloy Mar 15 '13 at 20:51
• Apparently, the list has been updated. – David Mitra Jun 19 '13 at 16:15

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.

• Not really. From $S_n > T_n$ you can only conclude that $\lim S_n \ge \lim T_n$. – lhf Jul 10 '11 at 21:24
• @lhf: That's right, but that can be easily fixed here (with $S_n = 1 + 1/2 + \dots + 1/2n$ and $T_n = 1 + 1/2 + \dots + 1/n$): we can use a better inequality, like say $S_n \ge T_n + 1/2$ (using just the first step) to conclude that $\lim S_n \ge \lim T_n + 1/2$, contradicting $S = \lim S_n = \lim T_n$. – ShreevatsaR Jul 11 '11 at 4:18
• I think "better" would be to use only $\frac{1}{4}$ in the first line, then $S - \frac{1}{4} \geq S$ – dEmigOd Feb 14 at 7:35

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.

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.

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 $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n}$.

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.

• If you look at a Riemann sum for intervals with width 1, you can pretty quickly see that the integral of 1/x from 1 to infinity must be less than the sum of the harmonic series. – Isaac Jul 21 '10 at 5:51
• Thank you for adding this answer. I was hoping to avoid an answer that involved integration, so I also prefer AgCl's answer. But I am happy to see more than one demonstration/proof. – bryn Jul 22 '10 at 11:33
• The sum is closer to the integral from $\frac{1}{2}$ to $n+\frac{1}{2}$ of $\frac{1}{x}$, which is $log(2n+1)$ math.stackexchange.com/a/1602945/134791 – Jaume Oliver Lafont Jan 25 '16 at 23:02

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.

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 $y = \dfrac {1}{x}$:

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 $p>1$.

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

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

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.

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

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

using the law of Geometric series

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

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

• Hm, the lower bound goes to $-\infty$ it appears. – Simply Beautiful Art Oct 6 '16 at 16:40
• @SimpleArt The upper bound goes to 0 and The Lower goes to $+\infty$ ,,,$-\ln 0^+$ – mhd.math Oct 7 '16 at 11:20
• Oh right, duh, didn't quite use that FTOC correctly. – Simply Beautiful Art Oct 7 '16 at 13:19

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.

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

• Ideally use Latex. – Meow Nov 1 '13 at 17:13
• 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. – Lucian Jan 14 '15 at 16:32

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:

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".

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 $\log (1 + x)$.

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.

• use \log to get nice formatting for $\log$ – Tyler Nov 1 '13 at 17:24

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 $$\sum_{n=1}^\infty\frac1n = \lim_{n\to\infty}H_n = \infty.$$

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 $S > S$.

• O.o This. Is. Amazing!! =) – user378947 Dec 12 '16 at 1:18
• You can also see it here math.stackexchange.com/questions/1160527/… – user8795 Dec 12 '16 at 1:20
• I have saved this to my personal The Book :) That being said... Come on! The last inequality itself is proof enough!! :P – user378947 Dec 12 '16 at 1:31

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}

• But isn't the series for $\ln(1-x)$ only valid for $-1\le x<1$? – TheSimpliFire Mar 17 '18 at 9:31
• Yes, but since the limit as $x\to1$ in $x^n$ is monotone, it equals the asked series, if they exist. – Simply Beautiful Art Mar 19 '18 at 2:33

$$\frac{1}{2} < \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n} < 1 \;\;\;(1)$$

Since $$\displaystyle H_m = \sum_{n=1}^{m}\frac{1}{n}$$ is increasing, if it is bounded, it converges, otherwise diverges to infinity.

Assume it converges, then:

$$\sum_{n=1}^{\infty}\frac{1}{n} = 2\frac{1}{2}+2\frac{1}{4}+2\frac{1}{6}+...+\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n} = \sum_{n=1}^{\infty}\frac{1}{n} + \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}$$

which is giving

$$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}=0$$

and that is not the case.

For $$\displaystyle (1)$$, the left side is obvious, the right side

$$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n} = \sum_{n=1}^{\infty}\frac{1}{2n-1}-\frac{1}{2n} = \sum_{n=1}^{\infty}\frac{1}{4n^2-2n}$$ $$< \frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n^2}<\frac{1}{2} + \frac{1}{2}\sum_{n=2}^{\infty} \left (\frac{1}{n-1}-\frac{1}{n} \right )$$

$$= \frac{1}{2} + \frac{1}{2} + \frac{1}{2}\sum_{n=2}^{\infty} \left (-\frac{1}{n}+\frac{1}{n} \right ) = 1$$

by canceling.

A series converges if and only if the tail of the series tends to zero, i.e. the summation from N to infinity tends to zero for N to infinity. But in case of the Harmonic series, we have that the summation from N to 2 N is larger than the smallest term ( which is 1/(2N)), times the number of terms N, which yields 1/2. So, the tail clearly does not tend to zero for N to infinity.

• What do you think you were adding that hasn't been addressed thoroughly? – user223391 Jan 31 '16 at 1:05
• @avid19 I added an argument that's tl;dr-proof. – Count Iblis Jan 31 '16 at 4:29
• @ZacharySelk My proof is the best, as it's the most concise, self contained proof given. Unlike the other proofs my proof can be modified into an argument that a 6 year old could understand (with some effort). – Count Iblis Oct 6 '16 at 22:35
• @CountIblis Your proof is mathematically identical to the six-year-earlier (and top-voted, and accepted) answer by AgCl, which also explains the situation much more clearly. (Note that bringing tails into the picture is unnecessary, and just adds a layer of complexity: we can reason about the blocks directly as AgCl does, and as you do essentially, to show that the harmonic series goes to infinity.) – Noah Schweber Oct 11 '16 at 1:52