An Interesting Sum With Reciprocals

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[@stevemonkey6666]

Well, okay. But the whole solution assumes you know what the answer is in the first place.

[@qq3088]

It can be found without knowing the answer. The sum of many harmonic series can be found using the integral of a geometric series

[@misterdubity3073]

Would have been nice to see this worked in the opposite direction; start with the series, then add in the x terms, then differentiate .... find 1/(1+x^3) ... integrate that. When @0:52 you start with 1/(1+x^3), it's like Deus Ex Machina

[@SyberMath]

😮

[@johnbutler4631]

Very clever and counterintuitive approach

[@GeoffryGifari]

Huh... the first thing I see are numbers with interval 3 starting from 1. No clue 1 + x^3 fits that pattern

[@armacham]

Yes, I have no idea where "1/(1 + x^3)" comes from or how it relates to the original equation or how integrating it solve anything
I wrote it as "1/(6a+1) - 1/(6a + 4)" where a starts at zero and increases by 1 to infinity

[@bernhardbauer5301]

What is it this guy tries to sum?

[@doctorb9264]

Interesting approach. I wonder of there is another method ?

[@meurdesoifphilippe5405]

It's very common to calculate the alternated sum of reciprocals of odds (=pi/4) by integrating 1/(x^2+1). This one is a similar approach. It can be easily generalized to calculate sum (-1)^k/(nk+r) with int 0 to 1 of x^(r-1)/(x^n+1).

[@Beathan64]

Why there was not possible to calculate it as the difference between two infinite sums (sequences)?
1+ (1/7+1/13... ) - (1/4+1/10...)

[@SyberMath]

I think those series do not converge.

[@vladimirkaplun5774]

This one is real stuff

[@e4uclassrooms725]

❤❤🎉🎉

[@yoav613]

Very nice😊💯

[@SyberMath]

Thanks ✌

[@VanDerHaegenTheStampede]

We can derive general formulas for the integral of *f(x)=(x^b)/(1+x^a)* as *alternating series:*
- The Integral *∫f(x)dx over [0, 1]* can be expressed as *∑ [ (-1)^k / (k*a+b+1) , {k, 0, ∞} ].*
- Also, for the improper integral *∫f(x)dx over [1, ∞)* , we get *∑ [ (-1)^k / ( (k+1)*a-b-1) , {k, 0, ∞} ].*
- Finally, the improper integral *∫f(x)dx over [0, ∞)* can be expressed as *∑ [ (-1)^k / (k*a+b+1) - (-1)^k / ( (k+1)*a-b-1) , {k, 0, ∞} ].*

[@paulortega5317]

Terrific approach

[@SyberMath]

Thanks!

[@pranjalsrivastava4260]

How can you put x=1 at the end?
At x=1 the original series is 1-1+1-1+......=1/2 is nonsense
x tends to 1from the left side, for big exponents, say (1-)^100/100, here the numerator itself tend to zero and is not equal to 1

[@scottleung9587]

Cool!

[@zakariakhalifa9681]

From 6:20 to 6:30 I didn't understand where artan came from.?
Please give me a hint

[@SyberMath]

arctan comes from the integral of du/(u^2+a^2)

[@zakariakhalifa9681]

@@SyberMath now I got you
Thanks

[@surelyred]

nice

[@SyberMath]

Thanks

[@ismaelperbech]

You can use a very dirty trick, take the series ln(x+1) = x - (x^2)/2 + (x^3)/3 - ...
Dividing by x bouth sides you reach (ln(x+1))/x = 1 - (x/2) + (x^2)/3 - (x^3)/4 + ...
If you evaluate ln(x + 1)/x with wx, (w^2)x and (w^3)x where w is the first cube root of unity and take the mean:
(1/3)*(ln(x+1)/x + ln(wx + 1)/(wx) + (ln((w^2)x + 1))/((w^2)x)) = 1 - (x^3)/4 + (x^6)/7 - (x^9)/10 + ...
This is because if you take the power of the 3 cube roots of unity which is not multiple of 3, the roots just permute and the sum will cancel, but if you take a power which is multiple of 3, all of the roots collapses to 1, and the sum is 3. In this way you can filter just the sum of just the powers multiple of 3. We just need to evaluate in 1 to get our result. Let S = 1 - 1/4 + 1/7 -....
S = (1/3)*(ln(2)/1 + ln(w+1)/w + ln(w^2 + 1)/(w^2))
Notice that w^2 + 1 = 1/(w + 1)
so ln(w^2 + 1) = ln(1/(w+1)) = -ln(w+1)
So S = (1/3)(ln(2) + ln(w+1)(1/w - 1/(w^2))
1/w - 1/w^2 = (w - 1)/w^2
Notice that w = 1/w^2 so the result is (w-1)w = w^2 - w
First w + 1 is the first sixth root of unity which is e^(2π/6)i, so ln(w+1) = (2π/6)i
Second w = -1/2 + ((√3)/2)i and w^2 = -1/2 - ((√3)/2)i so the difference w^2 - w = (-√3)i
Finally S = (1/3)(ln(2) + (2π/6)i(-√3)i = (1/3)(ln(2) + (π/3)√3) = ln(2)/3 + (π√3)/9
It is dirty because of convergence of the series of the logarithm as well as applying the rules of the logarithm when they could be violated when evaluating a logarithm of a complex number. Leave your thoughts😊

[@elinope4745]

I don't understand, I envy your math skills

[@SyberMath]

Wow! Mind-blowing 😍

[@kazebaret]

What you wrote is more than Math, is pure Art! Well done!

[@FisicTrapella]

I just set the general term for the sum as
S = (-1)^(2n+1) · [1 / (3n-2)]
where n goes from 1 to infinity... But I failed to evaluate it...🤫

[@peterromero284]

I lost the plot before the first minute was up

[@bosorot]

as hard as Basel problem

[@juanuribebravo6196]

Primer like, primer comentario