I challenge you plebs to prove this infinite sum

Surely this is bait. Have you never heard of proof by contradiction?

Pfff it is not that hard, introduce Bernstein polynoms and use the Parseval formula.

>introduce Bernstein polynoms
Surely you mean BerenSTAIN?

limit Veeky Forums -> infinity = /b/

i remember doing this one in high school

No I'm just incredibly retarded : I meaned Bernouilli's numbers :

[math]
B_0(X) = 1
[/math]

[math]
B'_{n+1}(X) = (n+1)B_n(x) [/math]
[math]
\int_{[0,1]} B_n(t)dt

[/math]

defines a sequence of polynoms, and denote [math] b_n = B_n(0) [/math]

Then you have the Bernouilli number sequence. Using the FOurier developpment of [math]B_n(X) [/math] on [math] [0,1] [/math] and the Parseval formula, you can compute :

[eqn]
\sum_{ n \geq 1} \frac{1}{n^{2k} }
[/eqn]

How the fuck would you solve this shit using Fourier series, I can't seem to find a function that will let me solve this fucking sum

I remember proving that the sum over 1/n^2 is pi^2/6 using the Fourier series of |x|
So i guess that would be a starting point

Wait, so it doesn't involve that retarded sine formula nobody bothers proving?

...

We know that [math] \sum_{n=1}^{\infty} 1/n^2=\pi^2/6[/math]. So
[math] \sum_{n=1}^{\infty} 1/n^6=(\sum_{n=1}^{\infty} 1/n^2)^3=\pi^6/6^3=\pi^6/945[/math].

Is math a meme?

how do you do formulas in Veeky Forums?

I am a Calc 2 Babby and to be quite honest I am #triggered at the fact that for most sums we can at best prove they converge but have no fucking idea to what.

When will this pain end? Will numerical analysis save me from the nothing I've become next year or will I have to wait until my junior year so that real analysis teach me how not to be a faggot?

Or is real analysis not even enough? Fuck, do I need a PhD in infinite series or what?

...

Use the TEX icon.

I guess it's more [math] x \mapsto \pi - x [/math] on [math] [0,2\pi] [/math]
For the other values, you can or use cotan, or the Bernouilli Polynoms.

Any continuous function can be developped as a Fourier series on any compact.

?_?

Formula using Bernoulli numbers ? You mean Euler-Maclaurin formula ?

Lol.

Most of differential equations can not be solved.

Most real numbers are not "computable".

We can even not find the roots of polynoms, or compute power series.

Maths is a hard world. Deal with it.

If this is all true then tbqh I think it is a time for a new axiom.

Consider the function [math]\displaystyle f(z) = \frac{2\pi i}{z^6(e^{2\pi i z} - 1)}[/math]. f has a pole of order 1 at each nonzero integer n with residue [math]\frac{1}{n^6}[/math].
Now, consider the square [math]K_N = \mathrm{conv}\left\{\left(N+\frac{1}{2}\right)(\epsilon_1 + i \epsilon_2), (\epsilon_1, \epsilon_2)\in \{-1,1\}^2\right\}[/math].
The boundary [math]\partial K_N[/math] does not contain any pole of f, hence by the residue theorem [math]\displaystyle \int_{\partial K_N} f(z)dz = \sum_{n = -N}^N Res_n(f) = Res_0(f) + 2\sum_{n=1}^N \frac{1}{n^6}[/math].
Now, it is easily seen that [math]\int_{\partial K_N} f(z)dz = O\left(\frac{1}{N^5}\right)[/math].
Finally, we get [math]\displaystyle \sum_{n = 1}^N \frac{1}{n} = -\frac{1}{2} Res_0(f) + O(1/N)[/math]. Passing to the limit, we get [math]\displaystyle \sum_{n \ge 1} \frac{1}{n^6} = -\frac{1}{2}Res_0(f)[/math]
Now, we have by definition of Bernoulli numbers, for z sufficiently small [math]\displaystyle \frac{2\pi iz}{e^{2\pi i z} - 1} = \sum_{n \ge 0} \frac{B_n}{n!}(2\pi i z)^n[/math]
Now, for z sufficiently small and nonzero, we get [math]\displaystyle f(z) = \sum_{n \ge 0} \frac{B_n}{n!}(2\pi i)^n z^{n-7}[/math] and hence [math]\displaystyle Res_0(f) = \frac{B_6}{6!} (2\pi i)^6[/math] and hence [math]\displaystyle \sum_{n \ge 1}\frac{1}{n^6} = \frac{B_62^5\pi^6}{720}[/math]
Now let's compute B_6: Remember the definining formal series equality [math]\displaystyle X = (e^X-1)\sum_{n \ge 0} \frac{B_n}{n!} X^n = \sum_{n \ge 1} \left(\sum_{k=1}^n {n \choose k} B_{n-k}\right)\frac{X^n}{n!}[/math]
We then get [math]B_0 = 1[/math] and [math]\sum_{k = 1}^n {n \choose k} B_{n-k} = 0[/math] for each [math] n \ge 2[/math]. Plugging in, we get [math]B_6 = 1/42[/math] and [math]\displaystyle \sum_{n \ge 1}\frac{1}{n^6} = \frac{\pi^6}{945}[/math]

Unnecessarily complicated.

Go ahead and provide a simple one if you can - that one is pretty nice.

Calculate the Fourier (sine) series of x, x^2, and x^3. This pretty much just means calculate 3 simple integrals.

Use basic algebraic manipulation to get the result of for the 1/n^2 series from the Fourier series of x.

Use this result + the Fourier series of x^2 to get the sum of the series 1/n^4.

Use this result + the Fourier series of x^3 to get the sum of the series 1/n^6.

That proof would be no simpler than the previous one - it just involves a different method. You not being familiar with the basics of complex analysis, or Bernoulli numbers, does not mean the proof is "unnecessarily complicated."

do it

I did it on my HW like 2 weeks ago.

I'm trying to do it but I get nonsense, I'm trying to do it fourier series method with pasrval

[eqn]\int sin(x^3)[/eqn]
help

It's not difficult, it's just a bunch of computation.
You want to use Parseval's theorem, which is
[eqn]
\frac{1}{2}a_0^2 + \sum_{n\geq1}(a_n^2 + b_n^2) = \frac{1}{\pi} \int_{-\pi}^{\pi} (f(x))^2 \text{d}x
[/eqn]
If you haven't worked with Fourier series before, you're going to want to look up how odd/even functions are evaluated.
While the Fourier series for [math]x[/math] and [math]x^2[/math] are simple to use, [math]x^3[/math] is a little more of a pain in the ass.

Follow this pattern.

- Calculate coefficients and you find...
[math]x = \sum\limits_{n = 1}^\infty {\left( {\frac{{2\ell {{\left( { - 1} \right)}^{n + 1}}}}{{\pi n}}} \right)} \sin \left( {n\pi x/\ell } \right)[/math]

- Apply Parseval
[math]\left\| x \right\|_{{L^2}}^2 = \sum\limits_{n = 1}^\infty {{{\left| {\frac{{2\ell {{\left( { - 1} \right)}^{n + 1}}}}{{\pi n}}} \right|}^2}} \left\| {\sin \left( {n\pi x/\ell } \right)} \right\|_{{L^2}}^2[/math]

=>

[math]\int\limits_0^\ell {{x^2}\operatorname{dx} } = \sum\limits_{n = 1}^\infty {{{\left| {\frac{{2\ell }}{{\pi n}}} \right|}^2}} \int\limits_0^\ell {{{\sin }^2}\left( {n\pi x/\ell } \right)} \operatorname{dx} [/math]

=>

[math]\frac{{{\ell ^3}}}{3} = \sum\limits_{n = 1}^\infty {{{\left| {\frac{{2\ell }}{{\pi n}}} \right|}^2}} \frac{\ell }{2}[/math]

=>

[math]\frac{{{\pi ^2}}}{6} = \sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} [/math]

Kek

It looks complicated but it is very robust (and actually involves very little computation). You just have to replace 6 with 2k everywhere to compute zeta(2k), there is literally nothing to change.
Besides, the method adapts readily to other series: If you have a sum of the form [math] \sum_{n\ge 1} \frac{1}{g(n)}[/math] with f an entire function, then you can try doing this with the function [math]\displaystyle f(z) = \frac{2\pi i}{g(z)(e^{2\pi i z} - 1}[/math] (maybe the contour will have to be more complicated though)

Top kek

The answer: Fourier analysis + Parseval's identity.

Your answer is logically incoherent because it assumes existence of dark numbers greater than 10^200 and that meta number like π can be treated as regular number

Speaking of infinite series
the limit as n goes infinity of (2^n +1)/(2^(n+1))

I'hops rule doesn't seem to be helping, so I have no idea how to determine that the limit is 1/2.

It might be as simple as the power of 2^(∞+1) expands greater than 2^∞. the Wolfram engine disagrees as it claims that both are still ∞ and that it would lead to ∞/∞

>tfw this brainlet can't grasp the mechanics of infinity

Replace (2^n +1) with (2^n) because that constant becomes insignificant. (2^n)/(2^(n+1)) = 1/2 for all n

That should have been obvious to me.

an intuitive way could have been splitting it into (2^n)/(2^(n+1)) + 1/(2^(n+1))

do the limit thing then 1/2 + 0 = 1/2