Is this series convergent or divergent?
Is this series convergent or divergent?
Justin Robinson
Other urls found in this thread:
wolframalpha.com
twitter.com
Blake Young
Convergent
Ethan Anderson
lim abs(a_n+1 / a_n) goes to zero so it converges
Luis Russell
James Lopez
yes it is
Brayden Howard
You can prove that [math] \sin \left( \frac{1}{2^n} \right) < \frac{1}{2^n} \implies n^2 \sin \left( \frac{1}{2^n} \right) < n^2 \frac{1}{2^n} \implies \sum n^2 \sin \left( \frac{1}{2^n} \right) < \sum n^2 \frac{1}{2^n} [/math]
Joshua Butler
...
Damn...
[math] \sin \left( \frac{1}{2^n} \right) = \frac{1}{2^n} + o \left( \frac{1}{2^n} \right) [/math]
Cameron White
so, we solve random homework now?
Nathaniel Powell
how do you prove the first statement
Carter Evans
You brainlet, every engineer on Veeky Forums knows that sinx=x.
Nicholas Parker
YEAH DUDE ESPECIALLY FOR BIG VALUES LMAO
James Gonzalez
[math]a^2-b^2 = (a+b)(a-b)[/math]
This is probably known to most brainlets.
but now observe:
[math]a-b = (\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})[/math]
and also:
[math]\sqrt[2^n]{a}-\sqrt[2^n]{b} = (\sqrt[2^{n+1}]{a}+\sqrt[2^{n+1}]{b})(\sqrt[2^{n+1}]{a}-\sqrt[2^{n+1}]{b})[/math]
therefore, by induction:
[math]a-b = \left[\prod\limits_{i=1}^\infty (\sqrt[2^i]{a}+\sqrt[2^i]{b})\right]\times \lim\limits_{i\rightarrow \infty}(\sqrt[2^i]{a}-\sqrt[2^i]{b})[/math]
But now absurdity is at hand!
because :
[math]\lim\limits_{i\rightarrow \infty}(\sqrt[2^i]{a}-\sqrt[2^i]{b}) = 0[/math]
therefore we have
[math]a-b = \left[\prod\limits_{i=1}^\infty (\sqrt[2^i]{a}+\sqrt[2^i]{b})\right]\times 0 = 0 [/math]
Hence
[math] a = b \quad \forall a,b\in \mathbb{R}[/math]
[math]\mathbb{Y} \mathbb{O}\mathbb{U}\mathbb{M}\mathbb{A}\mathbb{D}?[/math]
Mason Evans
Plonk it into your calculator using the sequence function and find out yourself.
Alexander Harris
Programmer spotted :D
Ian Wright
Let [math] f(x) = sin(x) - x [/math] Our goal is to prove that for [math] x > 0, f(x) < 0 [/math]. We can do this using elementary calculus. First notice that because sine is bounded by 1 this inequality is always true for x>1. So we actually only need to prove it for [math] 0 < x < 1 [/math].
First notice that [math] f(0) = sin(0) + 0 = 0 + 0 = 0 [/math]. Now take the derivative [math] f'(x) = cos(x) - 1 [/math]. In the interval [math] 0 < x < 1, cos(1) < cos(x) < 1 [/math] therefore [math] cos(x) - 1 < 0 [/math] in this interval. As the derivative is always negative in this interval, [math] f(x) [/math] is decreasing, and it is decreasing from 0. Therefore it is always negative. Proving that indeed [math] f(x) < 0 [/math].
Daily reminder that you learn calculus to be able to prove these kinds of inequalities without much effort.
Gabriel Phillips
no, [math]\lim\limits_{n\to\infty} f(n)g(n) = \left(\lim\limits_{n\to\infty} f(n)\right)\cdot\left(\lim\limits_{n\to\infty} g(n)\right)[/math] if both smaller limits exist.
so while [math]a-b = \lim\limits_{n\to\infty} \left[\prod\limits_{k=1}^n\left(\sqrt[2^k]{a}+\sqrt[2^k]{b}\right)\right]\cdot \left(\sqrt[2^k]{a}-\sqrt[2^k]{b}\right)[/math],
this can't be split into two limits since [math]\prod\limits_{i=k}^\infty \left(\sqrt[2^k]{a}+\sqrt[2^k]{b}\right) [/math] is divergent, as the factors approach 2.
new take on classic bullshit math.
Evan Martinez
Its convergent. Just plugged it into Desmos.
Wyatt Campbell
oh the horror, lets go back to ice/lava suns and iq whining
Easton Collins
>absurdity is at hand!
thou sayest it
Mark 15:2
Angel Lee
convergent, duh
sin(0.5^n)
Kevin Carter
>1 = n*(1/n)
>lim n-> infinity n*(1/n) = (lim n-> infinity n)*0 = 0
>infinity*0 = 0
>hurr but now absurdity is at hand
why do brainlets like you exist?
Ryder Allen
I GOT YOU NOW