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 Nov25 asked Proving that if $a_n\in\mathbb{Z}$ for all $n$ then it's limit is also in $\mathbb{Z}$? Nov21 comment Proving $0$ is the limit of $\frac{n}{2^n-1}$ from definition alone Well, the second way is pretty similar to Pambos's idea, but I really don't feel comfortable with logs. Is it something you expect I'd be introduced to or should I make an effort learn it on my own? Nov21 accepted Proving $0$ is the limit of $\frac{n}{2^n-1}$ from definition alone Nov21 comment Proving a vector space over itself have no subspaces In the end, after actually understanding how to use the dimensions, I found the other way simpler. But thanks for the help! Nov21 accepted Proving a vector space over itself have no subspaces Nov20 comment Proving $0$ is the limit of $\frac{n}{2^n-1}$ from definition alone Both answers say to use logs, but as of now we never had to use stuff that wasn't thought (or at least shown) in the course apart from basic algebra. Do you think logs enter this definition? Since I most definitly remeber nothing about them from high school... Nov20 asked Proving $0$ is the limit of $\frac{n}{2^n-1}$ from definition alone Nov19 comment Proving a vector space over itself have no subspaces Thanks for the edits! English is not my mother language so translating is sometimes difficult for me. Nov19 comment Proving a vector space over itself have no subspaces Ok, that did help me. I now know that if $1_F\in U$ it can't be a vector subspace. But I'm not sure what happens if $1_F\notin U$, since the axioms for vector spaces don't require a vector identity element, right? Nov19 asked Proving a vector space over itself have no subspaces Nov14 accepted Proving the sup of \$\{\frac{n-m}{n+m}|n,m\in \mathbb{N}, m