# Jonas Meyer

Dubuque, IA

 2d awarded Good Answer Dec 1 comment Banning a certain user$\text{@}$AhaanRungta: Thanks! Dec 1 comment Banning a certain userThanks. What's a multiple account? Dec 1 comment Banning a certain userWhat's a mutli? Nov 30 comment How to prove the closed form $\left(\frac {1}{1-x}\right)^2 = \sum_{n=0}^{\infty}(n+1)x^n$math.stackexchange.com/q/30732 Nov 30 comment Copying my solution, verbatim, is just so not cool@gnometorule: In the context of the sentence, the link is to an example of plagiarism between users, along with an editorial about how nonsensical such behavior is. That is always how I understood it, and still what it looks like. Nov 30 comment Dimension of diagonal matricesWhat was the result of your counting? Can you answer this when $n=1$, when $n=2$, or when $n=3$? Nov 30 answered when convergence in measure implies convergence almost surely Nov 30 comment Can't argue with success? Looking for "bad math" that "gets away with it"Depending on your perspective, often such use is not simply (shockingly) wrong, just lacking in proper justification and rigor. If it works each time, there is something to it. Nov 30 comment when convergence in measure implies convergence almost surelyDo you have a question? Nov 30 comment Seeking for construction s.t. every intersection contains at least 3 linesInteresting problem. I am curious, what led you to it? Although tagged (euclidean-geometry), it might help clarify the problem if you state explicitly that you are talking about lines in the Euclidean plane (if that is the case). Nov 30 comment Is it true that "$\lim_{n \rightarrow \infty} f(\frac{a}{n}) = 0$ implies $f(x)$ has a limit in $0$?(continued) @user83081: And if they don't, then all the terms are $0$. Nov 30 comment Is it true that "$\lim_{n \rightarrow \infty} f(\frac{a}{n}) = 0$ implies $f(x)$ has a limit in $0$?@user83081: (See also my comment on Ewan's answer.) To show that a given function $f$ satisfies the hypothesis means showing that if you start with arbitrary $a\in\mathbb R$, the sequence $f(a), f(a/2),f(a/3),\ldots$ converges to $0$. Pick any $a$ you want to, and it may or may not be the case that there exists some $n$ and some $k$ such that $1/e^k = a/n$ (again, there is no reason to assume $k=n$). Let your argument depend on whether or not such $n$ and $k$ exist. If they do, show that it happens only once, so all further terms in the sequence vanish. Nov 30 comment Is it true that "$\lim_{n \rightarrow \infty} f(\frac{a}{n}) = 0$ implies $f(x)$ has a limit in $0$?@user83081: In order to show that $f(a/n)$ converges to $0$ for each $a$, you must start with an arbitrary $a$, then consider the sequence $f(a), f(a/2), f(a/3),\ldots$. Ewan has given an example where you can give explicit reason why this sequence converges to $0$. You can break up your argument into $2$ cases: either $a$ is a rational multiple of $\dfrac{1}{\sqrt p}$ for some prime $p$, or it is not. The "$x$" appearing in the definition of the function is just for purposes of defining the function. Your question does not really make sense. Nov 30 comment Is it true that "$\lim_{n \rightarrow \infty} f(\frac{a}{n}) = 0$ implies $f(x)$ has a limit in $0$?@ParamanandSingh: Yes. Nov 30 comment Is it true that "$\lim_{n \rightarrow \infty} f(\frac{a}{n}) = 0$ implies $f(x)$ has a limit in $0$?@user83081: I wonder if you might be overloading $n$ there. Taken literally, of course that would be the case when $a=n/e^n$. But then $f(a/k)=0$ when $k>n$, so there is no problem. More to the point, for every $a$, there is at most one $k$ such that there exists $n$ such that $a/n=1/e^k$, and therefore this $n$ is also unique, and thus $f(a/m)=0$ when $m>n$. This uses the fact that no power of $e$ is an integer. Nov 30 comment Is it true that "$\lim_{n \rightarrow \infty} f(\frac{a}{n}) = 0$ implies $f(x)$ has a limit in $0$?Somewhat related: math.stackexchange.com/questions/63870/… Nov 30 answered Is it true that "$\lim_{n \rightarrow \infty} f(\frac{a}{n}) = 0$ implies $f(x)$ has a limit in $0$? Nov 30 comment Is it true that "$\lim_{n \rightarrow \infty} f(\frac{a}{n}) = 0$ implies $f(x)$ has a limit in $0$?By "$f$, has a limit in $0$" do you mean, $\lim\limits_{x\to 0}f(x) = 0$? If so, this should be stated. Nov 30 answered Group theoretic confusion on different group structures induced on the real line by ordinary multiplication and division