What are different ways to prove that $\sum_{n=1}^{\infty}\frac 1n$ is divergent?

Question

I have been studying Cauchy criterion for sequences, and have come across a rather simple proof for the harmonic series, and why it diverges.

More so, we have the following:

$$\sum_{n=1}^{\infty}\frac 1n=\infty\Rightarrow divergent$$

Here is my simple proof:

Consider the sequence $\left\{a_n\right\}_{n=1}^{\infty}$ such that $a_n=\frac 1n$, then $\forall \epsilon>0,\exists \ N \in\mathbb{R}$ for $m,n\in \mathbb{N}$, such that,

$$m,n>N\Rightarrow|a_m-a_n|<\epsilon$$

Pick $m=2n$, then we have the following:

$$|a_{2n}-a_n|=\sum_{k=n+1}^{2n}\frac {1}{k}\geq \sum_{k=n+1}^{2n} \frac {1}{2n}=\frac 12$$

So pick $\epsilon=\frac 12\Rightarrow \left\{a_n\right\}_{n=1}^{\infty}$ is not Cauchy and thus divergent.

My question is, are there any other slick and easy proofs for the above claim, and if so, what are they? This series at first surprised me, as it initially doesn't seem divergent.

Thanks in advance!

Answer 1

Here is a proof that uses the fact that if a sequence $a_n$ converges, then $a_{2n}$ converges to the same value.

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Let $S_n$ be the sequence given by $S_n=\sum_{k=n+1}^{2n}\frac{1}{k}$. Then, note that

$$S_{n+1}-S_n=\frac{1}{2n+2}+\frac1{2n+1}>0$$

Therefore, the sequence $S_n$ is increasing. Furthermore, $S_n$ satisfies the estimates

$$\frac12=S_n<S_n=\sum_{k=n+1}^{2n}\frac{1}{k}\le \sum_{k=n+1}^{2n}\frac{1}{n+1}=\frac{n}{n+1}<1$$

Therefore, inasmuch as $S_n$ is increasing and bounded above, $S_n$ converges.

Noting that $S_n=\sum_{k=1}^{2n}\frac1k -\sum_{k=1}^n\frac1k$ converges to a number between $1/2$ and $1$ (i.e., not $0$), we conclude that $\sum_{k=1}^\infty \frac1k$ must diverge.

Answer 2

We have $\lim_{x\to \infty}\ln x=\infty$ because $\ln x=\int_1^x(1/t)dt$ is monotonic and $\ln 3^n=n\ln 3>n$ for $n\in \mathbb N.$

For $y>0$ we have $\ln (1+y)|=\int_1^{1+y}(1/t)dt<\int_1^{1+y}1\cdot dt=y. $

Therefore $\sum_{j=2}^n(1/j)>\sum_{j=1}^n \ln (1+1/j)=\sum_{j=1}^n[\;\ln (j+1)-\ln j\;]=\ln (n+1),$ which $\to \infty$ as $n\to \infty.$

Answer 3

The slickest proof I have seen is the following.

We will argue by contradiction, assume that $\sum_{n=1}^{\infty}\dfrac{1}{n}=C <\infty.$ Then $\dfrac{C}{2}=\sum_{n=1}^{\infty}\dfrac{1}{2n}$, however

\begin{align} C=\sum_{n=1}^{\infty}\dfrac{1}{n} &=\sum_{n=1}^{\infty}\dfrac{1}{2n}+\sum_{n=0}^{\infty}\dfrac{1}{2n+1}\\ &>\sum_{n=1}^{\infty}\dfrac{1}{2n}+\sum_{n=0}^{\infty}\dfrac{1}{2n+2}\\ &=2\sum_{n=1}^{\infty}\dfrac{1}{2n} \end{align} Therefore, $$\dfrac{C}{2}>\sum_{n=1}^{\infty}\dfrac{1}{2n}$$ a contradiction.

Answer 4

The OP asks for different ways to prove that the harmonic series is divergent. Some were already provided.

Here's another one, which uses only a few ingredients from calculus (parametric integrals, geometric sum, expansion of $e^x$ close to $x = 0$)

Letting

$$h = \sum_{n=1}^ {\infty} n^{-1}\tag{1}$$

$$f(s) = \sum_{n=1}^{\infty} n^{-s}\tag{2}$$

then formally

$$h = \lim_{s\to 1} \, f(s)$$

Following Bernhard Riemann (1859, https://de.wikisource.org/wiki/%C3%9Cber_die_Anzahl_der_Primzahlen_unter_einer_gegebenen_Gr%C3%B6%C3%9Fe) we shall transform $f(s)$ into an integral the convergence or divergence of which can be checked easily.

Writing

$$n^{-s} = \frac{1}{c(s)} \int_{0}^{\infty} x^{s-1} e^{- n x} \,dx\tag{3}$$

with

$$c(s) = \int_{0}^{\infty} x^{s-1} e^{- x} \, dx$$

which are valid for $s \gt 1$ and $n \gt 0$.

Now doing the infinite sum under the integral (3) gives

$$\sum_{n=1}^{\infty} e^{- n x} = \frac{1}{e^x-1}$$

and the sum (2) becomes

$$fi(s) = \frac{1}{c(s)} \int_{0}^{\infty} x^{s-1} \frac{1}{e^x-1}\tag{4}$$

Now we attempt to let $s \to 1$.

The factor $c(s)$ obviously becomes 1.

The integral (4) may become divergent only close to $s = 0$ where the integrand behaves as $x^{s-2}$. Hence the integral is divergent (logarithmically) if $s = 1$ which in turn means divergence of the harmonic series $h$.

QED.