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 Oct5 awarded Popular Question Sep10 awarded Popular Question Jul2 awarded Curious Jan28 accepted Convergence of $\sum_{n=2}^\infty \frac{1}{n^\alpha \ln^\beta (n)}$ Jan28 comment Convergence of $\sum_{n=2}^\infty \frac{1}{n^\alpha \ln^\beta (n)}$ Sorry for commenting in an old post but I think that by the ratio test if $\alpha > 1$ the series converges. Dec31 accepted Every finite set contains its supremum: proof improvement. Dec31 accepted Every sequence is composed of isolated points? Dec31 accepted In a metric space, compactness implies completness Dec30 asked Convergence of $\sum_{n=2}^\infty \frac{1}{n^\alpha \ln^\beta (n)}$ Dec16 comment how to mathematically formulate the sign of a value? So $\mathrm{sgn}(a) :=\left\{\begin{array}{cc} \frac{|a|}{a} & \text{if } a \neq 0 \\ 0 & \text{if } a = 0\end{array}\right.$ Dec16 comment how to mathematically formulate the sign of a value? What about $a= 0$? Is there convention for $\mathrm{sgn}(0)$? Dec16 comment Every sequence is composed of isolated points? Corrected, thanks Dec16 asked Every sequence is composed of isolated points? Dec16 comment Every finite set contains its supremum: proof improvement. It is supposed that I can not conclude that $\max A = \sup A$ immediately, so I need to use supremum properties... Should I order elements of $A$ insted of using $\max A$? Dec16 awarded Custodian Dec16 reviewed Approve Every finite set contains its supremum: proof improvement. Dec16 asked Every finite set contains its supremum: proof improvement. Dec15 accepted In a metric space, if a set is compact, then it is closed: improving proof Dec15 revised In a metric space, if a set is compact, then it is closed: improving proof deleted 74 characters in body Dec15 comment In a metric space, if a set is compact, then it is closed: improving proof Thanks!, but I'm trying to understand this proof and unfortunately I've to rewrite it.