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This question already has an answer here:

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!

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marked as duplicate by Martin Sleziak, user99914, choco_addicted, JonMark Perry, yo' Apr 17 '17 at 16:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ $\sum_{k=n+1}^{2n}\frac1k \ne \frac12$. Why would you believe that there is equality here? $\endgroup$ – Mark Viola Apr 17 '17 at 2:50
  • $\begingroup$ @Dr.MV He means $\geq$. : $$\sum_{k=n+1}^{2n}\frac {1}{k}\geq n\sum_{k=1}^n \frac {1}{2n}$$ $\endgroup$ – N. S. Apr 17 '17 at 2:51
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    $\begingroup$ Your proof is basically just Cauchy Condensation Criteria. $\endgroup$ – N. S. Apr 17 '17 at 2:52
  • $\begingroup$ @N.S. Well that isn't even correct. Rather, we have $$\sum_{k=n+1}^{2n}\frac1k \ge \sum_{k=n+1}^{2n}\frac{1}{2n}=\frac12$$The final result is preserved, but there are flaws along the way. $\endgroup$ – Mark Viola Apr 17 '17 at 2:56
  • $\begingroup$ The Cauchy Condensation Test: If $a_n\geq a_{n+1}\geq 0$ for all $n$ then $\sum_na_n$ converges iff $\sum_n2^na_{(2^n)}$ converges.... If $a_n=1/n$ then $2^na_{(2^n)}=1.$ $\endgroup$ – DanielWainfleet Apr 17 '17 at 5:15
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Here is a proof that uses the fact that if a sequence $a_n$ converges, then $a_{2n}$ converges to the same value.


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.

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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.

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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 $n\to \infty.$

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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.

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    $\begingroup$ Now that's the most complicated proof to a 1st semester Calculus course problem I've ever seen. $\endgroup$ – yo' Apr 17 '17 at 16:33
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    $\begingroup$ That's ok, so you can learn a lot. Seriously, which parts do you consider complicated? The OP did not ask for an easy proof but for different one. This proof is by Leonard Euler from the 18th century, and I find it beautiful and instructive. It also forms the basis of Dirichlet's proof of the existence of infinitely many prime numbers in an arithmetic progression. $\endgroup$ – Dr. Wolfgang Hintze Apr 17 '17 at 20:19

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