If you really want a challenge, try summing 1/(n^6-1). For that, you might need to convert it into an integral. Interestingly, it also has 11/12 as part of the answer, but there's another term involving exponentials and pi.
@@tonyhaddad1394 My idea was to use partial fraction decomposition to write it as a sum of terms of the form 1/(n-(sixth root of unity)). Then using the fact that integral from 0 to 1 of t^(k-1)dt=1/k, we could rewrite each term as an integral. Then we can combine those using a geometric series. That trick works well for evaluating sums like 1-1/2+1/3-1/4+... and 1-1/3+1/5-1/7+... but now that I think about it, it probably won't work so well if the denominators are integers. Instead, you could use partial fractions to get it in the form A/(n-1)+B/(n+1)+(Cn+D)/(n^2+n+1)+(En+F)/(n^2-n+1). Then the terms with quadratics in the denominator can be handled using the fact that cot z=1/z+(sum from n=1 to infinity of 2z/(z^2-n^2*pi^2)) along with completing the square.
You can do this by taking f(z)= pi cot (pi z) /(z^6-1) and integrating around a large square in the complex plane. The sum of the residues is zero, and hence you get double series you want plus contributions from +1,-1 which are double poles, a contribution from zero and four contributions from the remaining sixth roots of unity (these give rise to terms like pi exp(i pi/3)cot(pi exp(i pi/3)). It's a bit of a mess to simplify in terms of hyperbolic trig functions, but should be okay.
In more detail: If n is an integer, not equal to 1 or -1, then Res( f(z), z=n)= 1/(n^6-1) We have Res( f(z), z=-1)=Res( f(z), z=1)= lim_{z to 1} (d/dz)[(z-1)²f(z)]= -5/12 And if w= exp(i pi/3), exp(-i pi/3), exp(2i pi/3) or exp(-2i pi/3) then Res( f(z),z=w)=lim_{z to w} (z-w)f(z)= (pi w/6) cot (pi w). Thus if S= sum_{n=2}^infty 1/(n^6-1) then 2S-1-2(5/12)+X=0 where X=pi/6[exp(i pi/3)cot( pi[1/2+isqrt(3)/2 )+exp(-i pi/3)cot( pi[1/2-isqrt(3)/2 ])+exp(2i pi/3)cot( pi[-1/2+isqrt(3)/2 ])+exp(2i pi/3)cot( pi[-1/2-isqrt(3)/2 ])] Now cot(pi[1/2+i sqrt(3)/2)=tan(-i sqrt(3)pi/2)=-i tanh(sqrt(3)pi/2) cot(pi[1/2-i sqrt(3)/2)=tan(i sqrt(3)pi/2)=i tanh(sqrt(3)pi/2) cot(pi[-1/2+i sqrt(3)/2)=i tanh(sqrt(3)pi/2) cot(pi[-1/2-i sqrt(3)/2)=-i tanh(sqrt(3)pi/2). Thus X=(4pi/6) sin(pi/3)tanh(sqrt(3)pi/2)=(pi/sqrt(3))tanh(sqrt(3)pi/2) and thus S=11/12-X/2=11/12-(pi/2sqrt(3))tanh(sqrt(3)pi/2)= 0.01759302638532157621375660449795520020
My first approach was to factor n^6-1 into (n-w)(n-w^2)(n-w^3)(n-w^4)(n-w^5)(n-1) for w = e^{\pi i /3} and go straight for a linear partial fraction decomposition - it got messy, and I made some mistakes, so I dialed it back and went for n^6-1 = (n-1)(n+1)(n^2-n+1)(n^2+n+1) and went for partial fractions with A/(n-1)+B/(n+1)+(Cn+D)/(n^2-n+1)+(En+F)/(n^2+n+1) - whence you can find A=-B=D=-F=1/2, C=E=0 - which makes a nice telescoping sum that leaves 1/2(1 + 1/2 + 1/3) = 11/12
I got that factorization via the intermediate factorization n^6-1 = (n^3-1)(n^3+1) (a difference of squares) and then further factoring with the well known sum and difference of cubes factorizations.
Si pero esas sumas parciales se le va mucho tiempo entre mas sean mas trabajo es por eso que es mejor hacerlo con polinomios de grado mayor para que sean menos fracciones.
Es darle vueltas a lo mismo porque si son factorizaciones del mismo polinomio pues con manipulación algebraica se puede llegar de uno a otro con la segunda factorización pues ya le salio directo
At 9:26 I thought when doing partial fraction decomposition with a non-factorable quadratic in the denominator that we needed to put a linear factor in the numerator rather than a constant. So shouldn't we use Cn + D and En + F instead of C and D for the (n^2 - n + 1) and (n^2 + n + 1) denominators?
Hi, 5:53 : to N, instead of to infinity, 6:11 : ok, correct, 9:11 : n⁴+n²+1=(n²-n+1)(n²+n+1) bravo! For fun: 17:47 : ok, nice (Côte d'Azur), 21:27 : nice photo composition with the splash screens.
@0:24 - since you are able to construct a closed form expression for the partial sum of the series, and the limit of the closed form (as big-N approaches infinity) exists, is it necessary to first prove that the series converges? In other words, does the existence of the limit of the closed form, as big-N approaches infinity, necessarily mean that the series itself converges? Or is that circular logic?
In your side calculations, your second set of partial fractions have quadratic denominators so surely your numerators should be linear. Cx+D and Ex+F rather than C and D?
Well, the (n^5+n^4+…+1) could be written as (n+1)(n^4+n^2+1), which would lead to the same result The same for the last factorization, where we could go further with the calculations
The ones I know are If x is even an k is a perfect square that is substracted you can use the difference of squares, if x is divisible by 3 and k is a cube you can use cubes addition or substraction, but i don't know any others without including complex roots
Here’s a quick way to solve simple (first degree polynomial) partial fractions decomposition: To get A, solve n sucht that the denominator equals 0, meaning n-1=0 => n=1, and replace the value of n into the full fraction, removing the denominator for A, meanning 1/(n+1)=1/2. For B, the denominator is n+1, so solving n+1=0 gives n=-1. Replacing it into the full fraction where the denominator part for B, n+1, is removed, you get 1/(n-1)=1/(-1-1)=-1/2. For second degree or more, it gets trickier and solving it the way it is in the videos seems to me the easiest one.
It's a slight logical leap but it's fine. Two lines (y = 1 and y = A(n+1) + B(n-1)) must intersect at no points, one point, or every point. They can't intersect at all but two points.
The answer would have been way more interesting, if the question was : find S = -1 + sum((n^4+n^3+n^2-n+1)/(n^6-1)). Since the latter gives -1/12 = Zeta(-1).
