*solves your equation with ease*

cum2soon
cum2soon

*solves your equation with ease*

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youtube.com/watch?v=b32hkQcM8U0

Dreamworx
Dreamworx

* Invents free Energy *

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Lord_Tryzalot
Lord_Tryzalot

Prove [math]\sum_{n\ \geqslant\ 1} \frac{\sin\left(2\, n\right)}{1 \,+\, \left(\cos\,n\right)^4}[/math] is convergent.

cum2soon
cum2soon

*never accomplishes anything past 19*

takes2long
takes2long

what's the motivation for this problem?

kizzmybutt
kizzmybutt

Making the difference between legit geniuses and autists spoonfed by their parents.

Emberfire
Emberfire

Int test [math]\sum\limits_{n\geq 1}\frac{\sin(2x)}{1+\cos(x)^4}\Rightarrow \int\limits_{0}^{\infty}\frac{\sin(2x)\,\mathrm{d}x}{1+\cos(x)^4}=\lim\limits_{t\to\infty} \int\limits_{0}^{t}\frac{2\sin(x)\cos(x)\,\mathrm{d}x}{1+\cos(x)^4}\quad \mathscr{Q.E.D.}[/math]

Flameblow
Flameblow

What a qt

TalkBomber
TalkBomber

[math]\int[/math]

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Spazyfool
Spazyfool

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Fried_Sushi
Fried_Sushi

its pretty fucking obvious that this series doesn't converge
for a series to converge, the absolute value of the term must approach 0 as n approaches infinity, but this obviously doesn't happen.
2*n arbitrarily approximates pi/2 + 2*pi*m for sufficiently large pairs of n and m. Take the expression 2*n - 2*pi*m. This expression can approximate pi/2 by applying the steps of the Euclidean algorithm, except, instead of using the standard termination condition, you terminate when the deviation from pi/2 is smaller than epsilon. The algorithm constructs the appropriate n and m when the termination condition is met, and the algorithm will always terminate for any arbitrarily small epsilon. So, we can always find an integer value of n such that 2*n is arbitrarily close to pi/2 + 2*pi*m, so sin(2*n) can be made arbitrarily close to 1 for infinitely many n. The denominator is bounded between 1 and 2, so, the general term does not approach 0 as n approaches infinity, so the series cannot converge.

Supergrass
Supergrass

...which is exactly what he proved

MPmaster
MPmaster

...which is exactly what he proved
Prove the series is convergent
...which is exactly what he proved
I mean, I know the facts are pretty obvious, but this seems like a question, and not an answer

askme
askme

The integral test doesn't work because the u substitution is completely fucked.
taking the cosine of t doesn't work because of its periodicity.

idontknow
idontknow

What is Jacob Barnett doing nowadays?

Spazyfool
Spazyfool

You also have sin(2*n) arbitrarily close to -1 for infinitely many n.

Clearly the terms don't go to zero, but it seems it may be possible to do a grouping of terms and have the sequence converge that way.

Raving_Cute
Raving_Cute

Dropped out of UNI and got into Red Devils. He's is trying to make it as a rapper now.

youtube.com/watch?v=b32hkQcM8U0

WebTool
WebTool

shitposting on /pol probably

CouchChiller
CouchChiller

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AwesomeTucker
AwesomeTucker

fuck off kurisu

CodeBuns
CodeBuns

time travel isn't possible

GoogleCat
GoogleCat

you literally proved yourself wrong within two weeks brainlet

Fuzzy_Logic
Fuzzy_Logic

he's cute and the same age as me. how can i meet up with him and do lewd things, Veeky Forums?

Nude_Bikergirl
Nude_Bikergirl

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