A cauchy sequence is bounded and $a_n=\sum^n_{k=1} \frac{1}{k}$ is unbounded. But because each term is smaller than the next it seems like I can find an $N$ such that the difference between all terms past $N$ is less than $\epsilon$ which would make it a cauchy sequence. Is it true that there is a sub sequence on $a_n$ right?
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Do you mean $\sum_{k=1}^n\frac1n$ or do you really mean $\sum_{k=1}^n \frac1k?$ – Thomas Andrews Sep 24 '19 at 22:17
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If you have shown it's not bounded you have effectively shown it's not a Cauchy sequence ($|a_n-a_m|$ for fixed $m$ will also be unbounded no matter how large $m$ is). – Winther Sep 24 '19 at 22:17
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Thanks thomas, I edited the post accordingly – Tsangares Sep 24 '19 at 22:19
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Never mind silly question. The terms get infinitely big so there is always a later term larger than $\epsilon$.
Tsangares
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