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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.

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marked as duplicate by Cm7F7Bb, Claude Leibovici, Najib Idrissi, Em., Watson Feb 19 '16 at 9:37

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  • $\begingroup$ Hint: $s_n\sim \log n$ asymptotically. $\endgroup$ – Math1000 Feb 19 '16 at 4:07
  • $\begingroup$ how are you choosing your $N$ ? $\endgroup$ – user300 Feb 19 '16 at 4:10
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    $\begingroup$ 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. $\endgroup$ – RRL Feb 19 '16 at 4:14
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    $\begingroup$ 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. $\endgroup$ – Steven Stadnicki Feb 19 '16 at 4:15
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    $\begingroup$ This looks rather similar to the proof that is given most frequently. See this question. $\endgroup$ – Martin Sleziak Feb 19 '16 at 6:05
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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 $$ \sum_{n=1}^{+\infty}2^na_{2^n}=\sum_{n=1}^{+\infty}\frac{2^n}{2^n}=+\infty. $$

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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.

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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: $$ H_n=\sum_{k=1}^{n}\frac{1}{k} \geq \log(n)+\frac{1}{2}. $$

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