>tfw you beat Wolfram Alpha
Is anyone here smart enough to work out the closed form? I'll post the answer after 20 replies
[math]\frac{1}{n}-\frac{1}{n+1}\sum_{1}^{n}\frac{1}{n}(-1)^n [/math]
0
>>7665075
also (-1)^(n+1)
>>7665054
>implying wolfram alpha is hard to beat
fucking. epic.
>>7665109
>/sci/ doesn't know what "closed form expression" means
>>7665054
5-pi^2/12-log(32)
yaaaaawn
>>7665054
shit I haven't done math in a while, this looks fun
wow you guys are pathetic, bunch of book regurgitators. The answer is (pi^2)/24 + ((ln2)^2)/2
>>7666187
this
>>7666187
If this is the correct answer, can someone please explain it in elaborate detail?
>>7666829
OP here unfortunately I lost the proof. I was trying to make a fractal by removing and adding squares from a square. I got this step like pattern that seemed to be infinite but when I computed the length of the line (that series) it converged so it obviously wasn't a fractal. I'll try and re-derive it but don't get your hopes up
>>7667392
lol no im not, stop shitposting
>>7665054
its a series within a series, I dont code so forgive the crappy notation:
sum(1/(n+1)+1/(n+2))*sum(1/(n+1)+1/(n+2))^n
>>7667408
>sum(1/(n+1)+1/(n+2))*sum(1/(n+1)+1/(n+2))^n
my bad, screwed up the addition, the actual series is:
sum(1/(n-1)+1/(n+2))*sum(1-1/(n+1)+(-1)^n/(n+2))^n
>>7666829
Not that guy, but here:
First write:
f(a,n) = sum(k=a->n, (-1)^(k-1) 1/k)
x = sum(n=1->inf, (1/(2n-1) - 1/(2n)) f(1,2n-1))
Expand the sum. Split into even and odd parts then recombine.
x = sum(n=1->inf, 1/(2n-1) f(1,2n-1)) - sum(n=1->inf, 1/(2n) f(1,2n-1))
x = sum(n=1->inf, 1/(2n-1) f(1,2n-1)) - sum(n=1->inf, 1/(2n) (f(1,2n) - (-1)^(2n-1) 1/(2n)))
x = sum(n=1->inf, 1/(2n-1) f(1,2n-1)) - sum(n=1->inf, 1/(2n) f(1,2n)) + sum(n=1->inf, (-1)^(2n-1) 1/(2n)^2)
x = sum(n=1->inf, (-1)^(n-1) 1/n f(1,n)) + sum(n=1->inf, (-1)^(2n-1) 1/(2n)^2)
Evaluate the part on the right:
R = sum(n=1->inf, (-1)^(2n-1) 1/(2n)^2)
R = -sum(n=1->inf, 1/(4n^2))
R = -1/4 sum(n=1->inf, 1/n^2)
R = -1/4 zeta(2)
And the part on the left:
L = sum(n=1->inf, (-1)^(n-1) 1/n f(1,n))
L = sum(n=1->inf, (-1)^(n-1) 1/n (f(1,inf) - f(1,n) + (-1)^(n-1) 1/n))
L = sum(n=1->inf, (-1)^(n-1) 1/n) f(1,inf) - sum(n=1->inf, (-1)^(n-1) 1/n f(1,n)) + sum(n=1->inf, ((-1)^(n-1) 1/n)^2)
L = sum(n=1->inf, (-1)^(n-1) 1/n)^2 - L + sum(n=1->inf, 1/n^2)
2L = log(2)^2 + zeta(2)
L = log(2)^2/2 + zeta(2)/2
Finally:
x = L + R
x = log(2)^2/2 + zeta(2)/2 - 1/4 zeta(2)
x = log(2)^2/2 + zeta(2)/4
x = log(2)^2/2 + pi^2/24
>>7667673
Can someone please display this is in Latex?
Three body equations...
Post some more problems/results like this, surprising ones with connections that aren't apparent at all.
>>7668201
Here you go:
[math]f(a,n) = sum(k=a->n, (-1)^(k-1) 1/k)
x = sum(n=1->inf, (1/(2n-1) - 1/(2n)) f(1,2n-1))
Expand the sum. Split into even and odd parts then recombine.
x = sum(n=1->inf, 1/(2n-1) f(1,2n-1)) - sum(n=1->inf, 1/(2n) f(1,2n-1))
x = sum(n=1->inf, 1/(2n-1) f(1,2n-1)) - sum(n=1->inf, 1/(2n) (f(1,2n) - (-1)^(2n-1) 1/(2n)))
x = sum(n=1->inf, 1/(2n-1) f(1,2n-1)) - sum(n=1->inf, 1/(2n) f(1,2n)) + sum(n=1->inf, (-1)^(2n-1) 1/(2n)^2)
x = sum(n=1->inf, (-1)^(n-1) 1/n f(1,n)) + sum(n=1->inf, (-1)^(2n-1) 1/(2n)^2)
Evaluate the part on the right:
R = sum(n=1->inf, (-1)^(2n-1) 1/(2n)^2)
R = -sum(n=1->inf, 1/(4n^2))
R = -1/4 sum(n=1->inf, 1/n^2)
R = -1/4 zeta(2)
And the part on the left:
L = sum(n=1->inf, (-1)^(n-1) 1/n f(1,n))
L = sum(n=1->inf, (-1)^(n-1) 1/n (f(1,inf) - f(1,n) + (-1)^(n-1) 1/n))
L = sum(n=1->inf, (-1)^(n-1) 1/n) f(1,inf) - sum(n=1->inf, (-1)^(n-1) 1/n f(1,n)) + sum(n=1->inf, ((-1)^(n-1) 1/n)^2)
L = sum(n=1->inf, (-1)^(n-1) 1/n)^2 - L + sum(n=1->inf, 1/n^2)
2L = log(2)^2 + zeta(2)
L = log(2)^2/2 + zeta(2)/2
Finally:
x = L + R
x = log(2)^2/2 + zeta(2)/2 - 1/4 zeta(2)
x = log(2)^2/2 + zeta(2)/4
x = log(2)^2/2 + pi^2/24[/math]
>>7669230
Nigga...
>>7669230
Faggot
>>7668201
>>7667673
------------------------------------------------------------------------
Not that guy but here:
First write:
[eqn]
f(n) = \sum_{k=1}^n \frac{(-1)^{k-1}}{k} \\
x = \sum_{n=1}^\infty\left( \frac{1}{2n-1}-\frac{1}{2n}\right) f(2n-1)
[/eqn]
Expand the sum. Split into even and odd parts then recombine.
[eqn]
x = \sum_{n=1}^\infty \frac{f(2n-1)}{2n-1} - \sum_{n=1}^\infty \frac{f(2n-1)}{2n} \\
x = \sum_{n=1}^\infty \frac{f(2n-1)}{2n-1} - \sum_{n=1}^\infty \frac{1}{2n}\left( f(2n) - \frac{(-1)^{2n-1}}{2n}\right) \\
x = \sum_{n=1}^\infty \frac{f(2n-1)}{2n-1} - \sum_{n=1}^\infty \frac{f(2n)}{2n} + \sum_{n=1}^\infty\frac{(-1)^{2n-1}}{(2n)^2} \\
x = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} f(n) + \sum_{n=1}^\infty \frac{(-1)^{2n-1}}{(2n)^2}
[/eqn]
Evaluate the part on the right:
[eqn]
R = \sum_{n=1}^\infty \frac{(-1)^{2n-1} }{(2n)^2} \\
R = -\sum_{n=1}^\infty \frac{1}{4n^2} \\
R = -\frac{1}{4} \sum_{n=1}^\infty \frac{1}{n^2} \\
R = \frac{-\zeta(2)}{4}
[/eqn]
And the part on the left:
[eqn]
L = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} f(n) \\
L = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\left( f(\infty) - f(n) + \frac{(-1)^{n-1}}{n}\right) \\
L = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} f(\infty) - \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} f(n) + \sum_{n=1}^\infty {\left(\frac{(-1)^{n-1}}{n}\right)}^2 \\
L = {\left(\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}\right)}^2 - L + \sum_{n=1}^\infty \frac{1}{n^2} \\
2L = log(2)^2 + \zeta(2) \\
L = \frac{log(2)^2}{2} + \frac{\zeta(2)}{2}
[/eqn]
Finally:
[eqn]
x = L + R \\
x = \frac{log(2)^2}{2} + \frac{\zeta(2)}{2}- \frac{\zeta(2)}{4} \\
x = \frac{log(2)^2}{2} + \frac{\zeta(2)}{4} \\
x = \frac{log(2)^2}{2} + \frac{\pi^2}{24}
[/eqn]------------------------------------------------------------------------
[Changed f(a,n) to f(n) as a was equal to 1 throughout]
>>7670558
It worked in the preview at least
