So, I stuck at a step to prove that $n^{1/10}$ is not a cauchy sequence. So, $$|y_n - y_m| = |n^{1/10} - m^{1/10}|$$
So, now how is the next step to show, that it is not a cauchy sequence?
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just apply the definition, what has to hold for $n,m,\varepsilon$? – user190080 Apr 25 '16 at 11:17
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e > 0, all n,m are natural numbers – Maxim Apr 25 '16 at 11:17
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1Assuming you're using the standard metric on the line, a sequence is Cauchy if and only if it converges. Does $\sqrt[10]{n}$ converge? – Umberto P. Apr 25 '16 at 11:18
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No not really, but I just need the correct mathematical prove – Maxim Apr 25 '16 at 11:19
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2Hint: Compute $|y_n-y_m|$ for $n=k^{10}$ and $m=(k+1)^{10}$, when $k\to\infty$. – Did Apr 25 '16 at 11:21
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1@Maxim You cannot prove that the sequence is Cauchy (under the standard metric), because it isn't to begin with. – egreg Apr 25 '16 at 11:28
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I´m sorry, I have falsely written. I want to show that is NOT a cauchy sequence. Sorry for the disturbing – Maxim Apr 25 '16 at 11:34
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If you want something direct, note that $x^{1/10}>2$ for $x>1024$, so $(2n)^{1/10}-n^{1/10}>n^{1/10}>1$ for all $n>1024$. – almagest Apr 25 '16 at 11:42
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2Cauchy $\implies$ bounded. – Martín-Blas Pérez Pinilla Apr 25 '16 at 11:54