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Show $$\sum_\limits{k=1}^\infty \frac 1 k$$ does not converge.

Attempt:

Let $s_n=\sum_\limits{k=1}^{n}1/k$, and let $\epsilon=1/2$. For all $N\in\mathbb{N}$, we have $$\left|s_{2n}-s_n\right|=\left|\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}\right|\geq1/2,\qquad\text{for all $n\geq N$}$$ Hence, $\{s_n\}_{n=1}^{\infty}$ is not a Cauchy sequence.

Since $\{s_n\}_{k=1}^{\infty}$ is not a Cauchy sequence which implies $\{s_n\}$ doesn't converge, we have $$\lim\limits_{n\rightarrow\infty}s_n=\lim\limits_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{1}{k}=\infty$$ Therefore, the infinite series $\sum_\limits{k=1}^{\infty}\frac{1}{k}$ does not converge.

I am not sure this is valid or not because I use contradiction to do these kind problem.

Martin Sleziak
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Simple
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  • Hint: $s_n\sim \log n$ asymptotically. – Math1000 Feb 19 '16 at 04:07
  • how are you choosing your $N$ ? – user300 Feb 19 '16 at 04:10
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    Your proof is valid as far as non-convergence is concerned. The sequence of partial sums fails the Cauchy criterion if there exists some $\epsilon> 0$ that for any $n$ there is some $m > n$ such that $|S_m - S_n| \geqslant \epsilon.$ You showed this by choosing $m = 2n$. This shows that the series is not convergent. To show the series must diverge to $+\infty$ , you only need to observe that the partial sums are increasing and cannot be bounded. – RRL Feb 19 '16 at 04:14
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    You should have stopped at '${s_n}$ is not a Cauchy sequence, which implies that it doesn't converge'; this already gives your conclusion (since ${s_n}$ is the sequence of sums you're studying). The next line (${s_n}$ doesn't converge implies that $\lim_n s_n=\infty$) is false. – Steven Stadnicki Feb 19 '16 at 04:15
  • Several variations of this problem have already been asked on this site. Reading the different answers will help you verify your own answer as they all share the same reasoning, even if subtly. For showing the divergence, just group terms in the series and see the pattern – Vishesh Feb 19 '16 at 04:21
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    This looks rather similar to the proof that is given most frequently. See this question. – Martin Sleziak Feb 19 '16 at 06:05

3 Answers3

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Your proof is fine, it just need a minor fix: since the sequence $\{s_n\}_{n\geq 1}$ $\color{red}{\text{is increasing}}$ and it is not a Cauchy sequence by the shown inequality, $\lim_{n\to +\infty}s_n = +\infty$.

You need the red part, since a sequence that is not a Cauchy sequence may also be an oscillating sequence, like in the case $a_n = \sin(n)$ or $a_n = 2+\cos(n)$.

The Cauchy condensation test is the usual way for proving that the harmonic series is divergent, but there are plenty of alternatives. For instance, since $f(x)=\frac{1}{x}$ is a convex function on $\mathbb{R}^+$, the Hermite-Hadamard inequality gives:

$$ \frac{1}{2}\left(\frac{1}{1}+\frac{1}{n}\right)+\left(\frac{1}{2}+\ldots+\frac{1}{n-1}\right)\geq \int_{1}^{n}\frac{dx}{x}$$ from which it follows that: $$ H_n=\sum_{k=1}^{n}\frac{1}{k} \geq \log(n)+\frac{1}{2}. $$

Jack D'Aurizio
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You can use Integral comparison test $$ \int_1^{+\infty}\frac1 x\,dx=\lim_{R\to+\infty}\log R=+\infty $$ or the Cauchy Condensation Critierion as well: $$ \sum_{n=1}^{+\infty}a_n<+\infty\; \Longleftrightarrow\;\sum_{n=1}^{+\infty}2^na_{2^n}<+\infty $$ where $a_n=1/n$; thus $$ \sum_{n=1}^{+\infty}2^na_{2^n}=\sum_{n=1}^{+\infty}\frac{2^n}{2^n}=+\infty. $$

Joe
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If it converged to some $L$, there would be some positive integer $k$ such that $\log(k)>L$, because the $\log(k)$ function is unbounded.

However, Lehmer's formula for the logarithm of positive integers $$\log(k)=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{k-1}+\frac{1-k}{k}+\frac{1}{k+1}+...+\frac{1}{2k-1}+\frac{1-k}{2k}+...$$ (see page 136 in http://matwbn.icm.edu.pl/ksiazki/aa/aa27/aa27121.pdf)

shows that you need to subtract a slowed-down version of the harmonic series from itself to form the logarithm of a positive integer. See this answer for numeric examples.

But you are never subtracting anything to your harmonic series, so there is no $\log(k)$ such that $\log(k)>L$ and no $L$ to which the series converges.

Community
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Jaume Oliver Lafont
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