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I have been told that if a series satisfies the Cauchy criterion, it converges. I was not quite convinced on this, however, since some of the posts on this website seemed to imply otherwise. I came up with what I thought was a counterexample: I argued that $1/n$ satisfies the Cauchy criterion but diverges. However, I've been told afterwards that $1/n$ doesn't satisfy the Cauchy criterion. Now, I've tried to show that the $1/n$ does not satisfy the Cauchy criterion but couldn't come up with a proof. Can someone provide me a proof that shows that the series $1/n$ does not satisfy the Cauchy criterion?

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    $\begingroup$ You must carefully distinguish between the sequence with term $\frac1n$, which converges, and the series with term $\frac1n$, the harmonic series, with partial sums $$H_n = \sum_{k=1}^n \frac1k,$$ which diverges. $\endgroup$ – Daniel Fischer Oct 13 '13 at 22:20
  • $\begingroup$ Sorry, I am referring to the harmonic series. The one that you posted. $\endgroup$ – CoffeeIsLife Oct 13 '13 at 22:24
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The sequence $\{1/n : n=1,2,3,\ldots\}$ satisfies the Cauchy criterion and converges.

The series $\displaystyle\sum_{n=1}^\infty \frac1n$ diverges. That is the same as saying that its sequence of partial sums $$ \left\{ \sum_{n=1}^m \frac1n : m = 1,2,3,\ldots \right\} $$ diverges. The latter sequence is not a Cauchy sequence. To see that it's not a Cauchy sequence without first considering whether it converges, consider the difference between the $m_1$th term and the $m_2$th term, whree $m_2>m_1$: \begin{align} & \phantom{={}} \left| \sum_{n=1}^{m_1} \frac1n - \sum_{n=1}^{m_2} \frac1n \right| \\[6pt] & =\left| \sum_{n=m_1+1}^{m_2} \frac1n \right| \ge \left| \int_{m_1+1}^{m_2} \frac1x\,dx \right| \\[6pt] & = \left|\log\frac{m_2}{m_1+1}\right|\to\infty\text{ as }m_2\to\infty. \end{align}

Don't confuse sequences with series.

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  • $\begingroup$ But why is the latter not a Cauchy sequence? I understand that it diverges but I want a proof that uses the Cauchy Criterion definition. $\endgroup$ – CoffeeIsLife Oct 13 '13 at 22:35
  • $\begingroup$ @user97554: You can prove in general that any sequence that diverges to $\infty$ must fail the Cauchy criterion. $\endgroup$ – Henning Makholm Oct 13 '13 at 22:44
  • $\begingroup$ I have found this math.stackexchange.com/questions/307330/… So I am not sure whether the inequality is valid...But the proof is true nevertheless. Thank you. $\endgroup$ – CoffeeIsLife Oct 13 '13 at 23:43
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    $\begingroup$ Think about $\displaystyle \int_5^6 \frac1x\,dx$. If $x$ is between $5$ and $6$, then $1/x$ is between $1/5$ and $1/6$, and is more than $1/6$. Therefore $\displaystyle \int_5^6 \frac1x\,dx>\int_5^6\frac16\,dx$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Oct 14 '13 at 17:45
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    $\begingroup$ @Resquiens : Say $4=m_1<m_2=8$. Then $$\sum_{n=1}^{m_1} \frac 1 n - \sum_{n=1}^{m_2} \frac 1 n =\sum_{n=1}^n \frac 1 n - \sum_{n=1}^8 \frac 1 n$$ $$=\left( \frac 1 1 + \frac 1 2 +\frac 1 3 + \frac 1 4 \right) - \left( \frac 1 1 + \frac 1 2 + \frac 1 3 + \frac 1 4 + \frac 1 5 + \frac 1 6 + \frac 1 7 + \frac 1 8 \right)$$ $$ = - \frac 1 5 - \frac 1 6 - \frac 1 7 - \frac 1 8 =-\sum_{n=5}^8 \frac 1 n = - \sum_{n=m_1+1}^{m_2} \frac 1 n.$$ Then take absolute values of both sides and the minus sign become plus signs ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 4 '15 at 1:26
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After you read Daniel's comment, look at $$s_n=\sum_{k=n+1}^{2n}\frac 1 k$$ and show this doesn't converge to $0$. In fact, it converges to $\log 2$.

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  • $\begingroup$ Could you elaborate? I am having trouble how it connects. $\endgroup$ – CoffeeIsLife Oct 13 '13 at 22:32
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    $\begingroup$ @user97554 You have $\lvert H_{2n} - H_n\rvert \geqslant \frac12$ for all $n$. So the series does not satisfy the Cauchy-criterion. $\endgroup$ – Daniel Fischer Oct 13 '13 at 22:44

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