Can someone please help me with this problem? I don't see

integral diverges, series diverges
it doesn't even tend to zero baka

>>8455612
The argument is not monotonicly decreasing you mong.

>>8455567
It doesn't converge:
for any N there exists some n>N such that |sin(2n)|>1/2 (I hope you know why). For this n |sin(2n)/(1+cos^4n)| > 1/4.

>>8455630
>for any N there exists some n>N
why say this instead of saying for all n?

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>>8455612
>>8455630
Uhh I think I trust him a little more than you guys

>>8455634
Okay, for all N.

>>8455636
small n -__-

>>8455635
We can't do nothing to the fact that sin(2n)/(1+cos^4n) doesn't even converge to zero, anon.

>>8455643
No, big N. It's a way to say "there exists a big term no matter how far you go"

>>8455647
Well he disagrees with you

https://www.youtube.com/watch?v=hBW4S9xcTOk

>>8455653
He doesn't disagree, he's being bullied by the show

>>8455664
But wait. Not like he's being bullied, but he just doesn't know integrals this well, he was probably expected to answer it's convergent right off the bat and therefore amaze the public. But it went wrong so they interrupted him loosing nothing (almost).

>>8455669
*know series
*divergent
I'm sleepy already

>>8455653
Faith. Hope. Charity.

>>8455653
"I can think in 4 and 5 dimensions."
What an insufferable faggot. Actually, every single person in that video is pure cancer except, maybe, the dad.

>>8455630
>|sin(2n)|>1/2
provide proof pls

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>its an american tv show acting like calc 2 is really difficult and complicated episode

12 is pretty young but its not like this stuff is difficult to learn. When you hear about actual intelligent people(not autistic """prodigies""") childhoods, they discover things on their own. Terrence Tao was collaberating with mathematicians at 8(!!!!). This kid is an autist, not a prodigy or even a savant.

>>8455769
>When you hear about actual intelligent people(not autistic """prodigies""") childhoods, they discover things on their own
Here you go: http://www.scribd.com/doc/233602815/Barnetts-Identity-Pdf1

Eat shit faggot.

>>8455752
A general fact is that [math]\{M, N+M, 2N+M, \ldots \}[/math] are dense mod 2pi for integer N and M.

Direct reasoning for this case:
[math]|\sin x|\leq \frac 1 2 \iff x \mod {\pi} \in [-\frac {\pi}{6}, \frac {\pi}{6}] [/math]

But since [math](\pi - \frac {\pi} {6}) - \frac {\pi}{6}=2.094.. > 2[/math] and [math]\frac {2 \pi}{6} < 2[/math] a segment [math][-\frac {\pi}{6} + 2, \frac{\pi}{6} + 2][/math] lies in [math](\frac{\pi}{6},\pi-\frac{\pi}{6})[/math]

Therefore whenever [math]|\sin x| \leq \frac 1 2[/math] it's also true that [math]|\sin (x + 2)| > \frac 1 2[/math].

>>8455780
*for natural N and M

>>8455772
>http://www.scribd.com/doc/233602815/Barnetts-Identity-Pdf1
Started reading. It's already the most fucked up thing I've ever read.

>>8455772
Lol, this was funny. Who wrote this?

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>>8455800

>>8455780
thanks famme. i was thinking in terms of irrationally of pi but I was unable to prove the denseness of integers mod pi.
the geometric approach makes it easier

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>>8455772
>http://www.scribd.com/doc/233602815/Barnetts-Identity-Pdf1

you do realize that's a joke right? as in, someone literally made that to meme, its not an actual math paper, just an elaborate /sci/ shitpost. my point stands

>>8455816
For a general case you first prove that a residue of [math]kN[/math] mod pi can be made arbitrary small:

Suppose [math]kN \mod \pi \in [-\varepsilon, \varepsilon][/math], since [math]\varepsilon \cdot \lfloor\ \frac {\pi}{\varepsilon}\rfloor \leq \pi \leq\varepsilon \cdot \lceil\ \frac {\pi}{\varepsilon} \rceil[/math] one of those sides is at least as close to [math]\pi[/math] as [math]\frac {\varepsilon} {2}[/math] you can multiply [math]kN[/math] by one of these rounded [math]\frac {\pi} {\varepsilon}[/math] (say it's [math]l[/math]) and get [math]klN \mod \pi \in [-\frac{\varepsilon}{2}, \frac{\varepsilon}{2}][/math]

This doesn't even use the nature of number pi nor number N, it was a simple general fact, but it solves our problem (and many similar), since pi is irrational we can get arbitrary small positive residue, and therefore all its multiples, therefore approach any number with any accuracy. So residues kN is a dense subset therefore kN + any number is dense as well.

>>8455567
Can someone help me figure out if this series converges?

[math] \sum_{n=1}^{\infty} {\frac 1 {n^3\sin^2n}}[/math]

I tried the integral test and ratio test already but neither helped :(

>>8455941
can't you just use a direct comparison with p-series?

>>8455981
nope
>>8455941
I doubt it converges. the sen is going to hit a number near zero that will fuck everything up at some point.

>>8455941
That's an interesting question
It has to do with rational approximations of [math]\pi[/math].
I mean it depends on whether there are infinite number of fast rational approximations like this for example:
[math]|\frac n m - \pi| < \frac {1} {n^3} \iff n \mod \pi < \frac {m}{n^2} \leq \frac {1}{n^{2}}[/math] in which case there is an infinite subsequence such that [math]sin^2(n)~\frac {1}{n^4}[/math] and therefore [math]\frac{1}{sin^2(n)n^3}~n[/math]

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>>8455941
>>8456029
I am not quite sure, correct me if I am wrong.
For the series to diverge, there has to be a sequence of natural numbers n, where n*pi is closer to a natural number than the previous member of the sequence.
In addition, the difference of the close natural number with n*pi has to converge faster or equal towards zero compared to the compressing poles.

>>8455567
>screenshot of a real life scene
rly made my think

>>8456366
Yeah, that's not enough to prove anything

Although Hurwitz's irrational number theorem says that there exist an infinite number of [math]p, q[/math] such that [math]|\pi - \frac p q| < \frac 1 {\sqrt{5} q^2}[/math] (and there's no better result for pi)
This yields
[math]p \mod \pi \in (-\frac 1 {q },\frac 1 {q })[/math] and since [math]p < 4q[/math]
[math]p \mod \pi \in (-\frac 4 {p},\frac 4 {p })[/math]
sin x < x for small x so:
[math]\frac {1}{p^3 \sin^2 p} > \frac{1}{16 p^{3-2}}= \frac{1}{16 p}[/math]
So it's true for infinitely many [math]p[/math] which is completely useless since they can occur too rare.

>>8456559
>>8455941

Ah, wait, this actually does prove this problem:
Whenever [math]p \mod \pi \in (-\frac {4}{p}, \frac{4}{p})[/math] it's also [math]\alpha = p \frac{\alpha}{p} \mod \pi \in (-\frac {4 \alpha}{p^2}, \frac {4 \alpha}{p^2})[/math] for any number [math]\alpha[/math]
So [math]\sum_p^{2p} \frac{1}{n^3 \sin^2(n)}\geq \sum_p^{2p} \frac{1}{16 n^3 \frac {n^2}{p^4}}=\sum_p^{2p} \frac{p^4}{16 n^5 } \geq \frac{p^4}{16} \int_{p-1}^{2p}\frac{dx}{x^5}\geq \frac{15}{16\cdot 64}>0[/math]
So there are infinitely many parts of series with at least this unzero sum.

So yeah. This series diverges

>>8456559
>[math] \alpha \mod \pi\in (-\frac{4\alpha}{p^2},\frac{4\alpha}{p^2})[/math] for any [math] \alpha [/math]

Gotta say I don't buy this desu.

We know that [math] -4/p<p-n\pi<4/p [/math] for some integer n. It follows that [math] -4\alpha/p^2<\alpha-(n\frac{\alpha} {p})\pi<4\alpha/p^2 [/math]. However, if [math] \alpha/p [/math] is not an integer, this does not imply that [math] \alpha \mod \pi\in (-\frac{4\alpha}{p^2},\frac{4\alpha}{p^2})[/math], because we're only allowed to mod out by integer multiples of pi.

You can already see the problem in a simpler example. We know that [math] 0\leq 10\mod 9\leq 2 [/math] but it is not the case that [math]0\leq 15\mod 9\leq 3=2*1.5 [/math].

In particular, your bound for sin(n) for n between p and 2p falls apart, because n/p is not an integer for each such n.

Interesting approach though!

>>8457899
Oops, you're right, that's true. Dunno how to solve it then. Perhaps it converges...