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 2d accepted Finding $\lim\limits_{n\rightarrow \infty}\sum\limits_{r=1}^{n}\frac{1}{T_r}$ given $\sum\limits_{r=1}^{n}T_r=\frac{n(n+1)(n+2)(n+3)}{8}$ 2d comment Finding $\lim\limits_{n\rightarrow \infty}\sum\limits_{r=1}^{n}\frac{1}{T_r}$ given $\sum\limits_{r=1}^{n}T_r=\frac{n(n+1)(n+2)(n+3)}{8}$ Its is unknown. 2d asked Finding $\lim\limits_{n\rightarrow \infty}\sum\limits_{r=1}^{n}\frac{1}{T_r}$ given $\sum\limits_{r=1}^{n}T_r=\frac{n(n+1)(n+2)(n+3)}{8}$ Feb27 awarded Notable Question Jan27 awarded Popular Question Jan10 awarded Yearling Dec18 accepted How to prove Linear Independence Nov27 comment How to prove Linear Independence If $x>0$ then $c=1$ and if $x<0$ then $c=-1$. Nov27 asked How to prove Linear Independence Oct10 comment Intersection of minimal seperators Intersection may be minimal for some other pair of vertices. Definition of minimal separator says that it is a set of vertices $S$ for which there exist non adjacent vertices $a$ , $b$ such that $a$ & $b$ belongs to distinct connected components after removing vertices in $S$, and no proper subset of $S$ is an $a,b$ separator. Oct10 comment Intersection of minimal seperators Here I am talking about two different minimal separators. In general we can find two different minimal separators with non empty intersection. Oct10 asked Intersection of minimal seperators Oct3 awarded Popular Question Sep24 asked K-tree and Partial k-tree Sep9 accepted Intuition behind Upper central series of Group Sep9 comment Find upper limit for this probability using Chebychev's inequality. Given distribution is uniform on (0,10), find the p.d.f of uniform distribution. Sep9 asked Intuition behind Upper central series of Group Aug27 comment Proving the dimension of basis of given subspace Let us take standard basis for $\mathbb{R}^5$ and consider two dimension subspace of $\mathbb{R}^5$ such that last two coordinates are zero and sum of first three is zero. then its has a basis such that none of them belongs to standard basis of $\mathbb{R}^5$. Aug26 accepted Subgroup problem Aug26 comment Subgroup problem Yes you are right. $b \in B$ I have edited the question