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Nov
13
comment Given $a_{1}=1, \ a_{n+1}=a_{n}+\frac{1}{a_{n}}$, find $\lim \limits_{n\to\infty}\frac{a_{n}}{n}$
testing the first few elements gives me the impression that $\frac{(a_n)^2}{n}$ is increasing, so I'll guess and say 3.
Nov
13
answered if $r\in (0,1), 1-r$ are the only sub limits of $a_{n}$ then $f(a_{n})$ converges when
Nov
13
asked if $r\in (0,1), 1-r$ are the only sub limits of $a_{n}$ then $f(a_{n})$ converges when
Nov
13
comment Given $a_{1}=1, \ a_{n+1}=a_{n}+\frac{1}{a_{n}}$, find $\lim \limits_{n\to\infty}\frac{a_{n}}{n}$
perhaps you have a hint for $\frac{(a_{n})^2}{n}$? the closest I got was to show that $\frac{1}{n}\leq \frac{(a_{n})^2}{n} \leq 3$ which implies that if $b_{n}=\frac{(a_{n})^2}{n}$ is monotonic then the limit exists. but still I haven't showed that it's monotonic and even if it is, how do I find its limit?
Nov
13
comment Given $a_{1}=1, \ a_{n+1}=a_{n}+\frac{1}{a_{n}}$, find $\lim \limits_{n\to\infty}\frac{a_{n}}{n}$
that's what I tried by showing $a_{n}\leq n$ but it wasn't tight enough.
Nov
12
accepted Given $a_{1}=1, \ a_{n+1}=a_{n}+\frac{1}{a_{n}}$, find $\lim \limits_{n\to\infty}\frac{a_{n}}{n}$
Nov
12
accepted Convergence of $a_{0} = 0, a_{n}=f(a_{n-1})$ when $|f'(x)|\leq \frac{5}{6}$
Nov
12
comment Given $a_{1}=1, \ a_{n+1}=a_{n}+\frac{1}{a_{n}}$, find $\lim \limits_{n\to\infty}\frac{a_{n}}{n}$
yes, I was hoping that $a_{n}/n \to 1$ and then it's easy, but guess not :)
Nov
12
comment Given $a_{1}=1, \ a_{n+1}=a_{n}+\frac{1}{a_{n}}$, find $\lim \limits_{n\to\infty}\frac{a_{n}}{n}$
actually the next question is to look at $(a_{n})^2 / n$ ;)
Nov
12
revised Given $a_{1}=1, \ a_{n+1}=a_{n}+\frac{1}{a_{n}}$, find $\lim \limits_{n\to\infty}\frac{a_{n}}{n}$
edited body
Nov
12
asked Given $a_{1}=1, \ a_{n+1}=a_{n}+\frac{1}{a_{n}}$, find $\lim \limits_{n\to\infty}\frac{a_{n}}{n}$
Nov
12
accepted If $(x_{k})\to L$ and $\forall x_{i}\in (x_{k})$, $x_{i}$ is a subsequential limit of $a_{n}$ then
Nov
12
asked If $(x_{k})\to L$ and $\forall x_{i}\in (x_{k})$, $x_{i}$ is a subsequential limit of $a_{n}$ then
Nov
12
comment Convergence of $a_{0} = 0, a_{n}=f(a_{n-1})$ when $|f'(x)|\leq \frac{5}{6}$
Thanks, I believe it helped. See my edit.
Nov
12
revised Convergence of $a_{0} = 0, a_{n}=f(a_{n-1})$ when $|f'(x)|\leq \frac{5}{6}$
deleted 10 characters in body; added 42 characters in body
Nov
12
awarded  Editor
Nov
12
revised Convergence of $a_{0} = 0, a_{n}=f(a_{n-1})$ when $|f'(x)|\leq \frac{5}{6}$
added 428 characters in body; deleted 5 characters in body; added 26 characters in body; added 9 characters in body
Nov
12
asked Convergence of $a_{0} = 0, a_{n}=f(a_{n-1})$ when $|f'(x)|\leq \frac{5}{6}$
Nov
12
awarded  Scholar
Nov
12
accepted sin(n) subsequence limits set