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Let $x_n=\sum_{j=1}^{n}1/j$. Show that for every $k$ we have $\lim_{n \to \infty}|x_{n+k}-x_n|=0$, yet $\{x_n\}$ is not Cauchy.

My attempt: $x_{n+k}-x_n = \sum_{j=n+1}^{n+k}1/j = \frac 1{n+1}+...+\frac 1{n+k}$. Then, $\lim_{n \to \infty} |\frac 1{n+1}+...+\frac 1{n+k}|=0$.

Next, prove that $\{x_n\}$ is not cauchy.

Suppose $x_n$ is cauchy, and let $S_i$ for $i\in Z^+$ be the partial sum of the series and $n+k=m$. Then, $\forall \varepsilon >0,$ there exists $M\in Z^+$ such that $|S_{m}-S_n|<\varepsilon$ for $ n\ge M$ and $m>n$.

Then, $|S_{m}-S_n|=|\sum_{j=n+1}^{m}1/j|<\varepsilon $ for all $n$ and $m\ge M$.

However, as $n$ fixed and $m \rightarrow \infty, |\sum_{j=n+1}^{m}1/j|$ does not converges since it is harmonic series, and this is contradiction.

Therefore, $x_n$ is not cauchy.

Is it okay? Could you pick mistakes?

Thank you in advance.

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  • $\begingroup$ I think you may want to rephrase the title statement that you are proving, to clarify that you are constructing a non-Cauchy $\{x_n\}$ such that for every fixed $k$ $lim_{n\rightarrow\infty}|x_{n+k}-x_n| = 0$. As opposed to the easy problem of constructing a non-Cauchy $\{x_n\}$ such that for some fixed $k$ $lim_{n\rightarrow\infty}|x_{n+k}-x_n| = 0$. The latter is trivial, e.g., $\{x_n\}=(0,1,0,1,0,1,...)$ and $k=2$. $\endgroup$
    – C Monsour
    Apr 17, 2018 at 11:48

1 Answer 1

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It's quite well done, but you cannot say that $\left(\sum_{j=n+1}^m\frac1j\right)_{m\in\mathbb N}$ is the harmonic series, because it isn't. What you can say is that, since the harmonic series diverges, if you remove from it finitely many terms, then you'll get a series which is divergent too.

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