*solves your equation with ease*

# *solves your equation with ease*

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* Invents free Energy *

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

*never accomplishes anything past 19*

what's the motivation for this problem?

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

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]

What a qt

>[math]\int[/math]

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.

...which is exactly what he proved

>...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

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.

What is Jacob Barnett doing nowadays?

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.

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

shitposting on /pol probably

fuck off kurisu

time travel isn't possible

you literally proved yourself wrong within two weeks brainlet

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