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seen Jul 23 '13 at 15:47

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
Nov
9
awarded  Supporter
Nov
8
comment sin(n) subsequence limits set
@Qiaochu: I'm afraid not, this sucks. Here's a more compact output, it only prints when n changes pastebin.com/raw.php?i=NnXpJFZP
Nov
8
awarded  Commentator
Nov
8
comment sin(n) subsequence limits set
@Qiaochu: there are certain values of n that repeat for many many epsilons
Nov
8
comment sin(n) subsequence limits set
@Qiaochu: like this pastebin.com/raw.php?i=aV0UbNUY ?
Nov
8
comment sin(n) subsequence limits set
@Qiaochu: you mean for smaller values of epsilon? This is how the test app currently looks like: pastebin.com/raw.php?i=YVX9fbc6
Nov
8
comment sin(n) subsequence limits set
@Qiaochu: here are some numbers pastebin.com/raw.php?i=pnhSL1rr - am I missing something obvious here?
Nov
8
comment sin(n) subsequence limits set
@Qiaochu: well as I predicted, the closer I want to get to an integer, the further I have to go in the naturals. But I have no idea how this helps, I mean it seems obvious that at some point I'll get close enough since I can go to infinity.