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.
 A: 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 $\log (1 + x)$.
A: 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: 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.
A: 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+.
A: 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
$$\sum_{n=1}^\infty\frac1n = \lim_{n\to\infty}H_n = \infty.$$
A: The proof I learned,  in Rosenlicht's Introduction to Analysis, published by Dover, is essentially a variant of the most popular answers above.
Namely,  we will show that the sequence $S_n=\sum_{k=1}^n \dfrac 1k$ is not Cauchy.
For given $N\gt0$, look at $\vert S_{2N}-S_N\vert=\dfrac1{2N}+\dots+\dfrac 1{N+1}\ge N\cdot \dfrac 1{2N}=\dfrac 12$.
A: 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.
A: 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.
A: 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 $\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n}$.
A: 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.
A: 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.
A: 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,
$$\begin{align}
\sum_{k=1}^\infty\frac1{k^s} & =\frac1{1-2^{1-s}}\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k^s} \\
\sum_{k=1}^\infty\frac1k & =\frac10\sum_{k=1}^\infty\frac{(-1)^{k+1}}k\tag{$s=1$} \\
& \to+\infty
\end{align}$$
A: One of the possible answers. 
First obviously:
$$ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n} > 0$$
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 would be giving
$$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}=0$$
and that is not the case.
A: Visualize the harmonic series  as the area of a sequence of rectangles:

$$ S:=1+\frac12+\frac13+\frac14+\frac15+\cdots$$
Now compute this same area using horizontal rectangles:

$$ 
\begin{align}
\textstyle S=\sum\text{width $\cdot$ height} &= 1\cdot\left(1-\frac12\right) + 2\cdot\left(\frac12-\frac13\right) + 3\cdot\left(\frac13-\frac14\right) + 4\cdot\left(\frac14-\frac15\right)+\cdots\\
&=\left(1-\frac12\right)+\left(1-\frac23\right)+\left(1-\frac34\right)+\left(1-\frac45\right)+\cdots\\
&=\frac12+\frac13+\frac14+\frac15+\cdots
\end{align}
$$
... and the first term in the sum has disappeared! Since $S=S-1$, the sum cannot be finite.

Here's the same argument as a one-liner:
$$
\sum_{n=1}^\infty\frac1n\stackrel{(a)}=
\sum_{n=1}^\infty\sum_{k=n}^\infty\left[\frac1k-\frac1{k+1}\right]
=\sum_{n=1}^\infty\sum_{k=n}^\infty\frac1{k(k+1)}
\stackrel{(b)}=
\sum_{k=1}^\infty\sum_{n=1}^k\frac1{k(k+1)}=\sum_{k=1}^\infty\frac1{k+1}
$$
In (a) we have a telescoping sum; in (b) the interchange in the order of summation is legal since all terms are nonnegative.
A: 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.
A: 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.
A: 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 $p>1$. 
A: 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.
A: 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.
A: $$\int_{0}^{\infty}e^{-nx}dx=\frac1n$$
$$\sum_{n=1}^{\infty}\int_{0}^{\infty}e^{-nx}dx=\lim_{ m \to \infty}\sum_{n=1}^{m}\frac1n$$
using the law of Geometric series
$$\int_{0}^{\infty}(\frac{1}{1-e^{-x}}-1)dx=\lim_{ m \to \infty}H_m$$
$$\lim_{ m \to \infty}H_m=\left [ \ln(e^x-1)-x \right ]_0^{\infty}\to\infty$$
A: 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.
A: 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 $S > S$.
A: \begin{align}
\sum_{n=1}^\infty\frac1n=\sum_{n=1}^\infty \int_0^1x^{n-1}\ dx=\int_0^1\sum_{n=1}^\infty x^{n-1}\ dx=\int_0^1\frac{dx}{1-x}=\int_0^1\frac{dx}{x}=\ln(1)-\ln(0)=\infty
\end{align}
A: This is based on the same idea as several other answers, but the presentation, I think, is sufficiently different to make it worth adding.  The key is to note that the inequalities $2\gt1$, $2/3\gt1/2$, and $2/5\gt1/3$ generalize to $2/(2n-1)\gt1/n$.
Let
$$S=1+{1\over2}+{1\over3}+{1\over4}+{1\over5}+{1\over6}+\cdots$$
If $S$ were finite then we would have
$$\begin{align}
2S&=2+{2\over2}+{2\over3}+{2\over4}+{2\over5}+{2\over6}+\cdots\\
&=\left(2+{2\over3}+{2\over5}+\cdots \right)+\left({2\over2}+{2\over4}+{2\over6}+\cdots \right)\\
&\gt\left(1+{1\over2}+{1\over3}+\cdots \right)+\left(1+{1\over2}+{1\over3}+\cdots \right)\\
&=2S
\end{align}$$
But the strict inequality $2S\gt2S$ is impossible. So $S$ cannot be finite.
A: Let $r=\frac{m}{n}$ be any positive rational number. Beginning with the equation
$$
r=\underbrace{\frac{1}{n}+\frac{1}{n}+\dots+\frac{1}{n}}_{m\text{ times}}
$$
and repeatedly applying the algebraic identity
$$
\frac{1}{p}=\frac{1}{p+1}+\frac{1}{p(p+1)}
$$
we can eventually find an Egyptian fraction representation
$$r=\frac{1}{n_1}+\frac{1}{n_2}+\dots+\frac{1}{n_k}$$
of $r$ with $n_1<n_2<\dots<n_k$. So $r<H_{n_k}$.
Since $r$ was arbitrary, it follows that the sequence of harmonic numbers is unbounded. Thus the harmonic series diverges.
(I think the greedy algorithm for constructing Egyptian fractions doesn't work here, as you already need to know that the harmonic series diverges in order to prove that the greedy algorithm works for arbitrarily large rationals...)
