my first alternating telescoping series

Alternating telescoping series 1/2-1/6+1/12-1/20+...
A good supplementary video: Evaluate infinite series by using power series: • 6 Infinite Series You ...
A good challenge: given 1+1/4+1/9+1/16+...=pi^2/6, what is the value of 1-1/4+1/9-1/16+...=?
(solution video here: • 1-1/4+1/9-1/16+... (z... )
The canvas print in the video: All series convergence tests: 👉 bit.ly/3GW8EJI
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[@gurkiratsingh7tha993]

This is an absolute masterpiece series

[@robertveith6383]

For a non-rigorous method: It equals
(1 - 1/2) - (1/2 - 1/3) + (1/3 - 1/4) - (1/4 - 1/5) + ... =
1 - 1/2 - 1/2 + 1/3 + 1/3 - 1/4 - 1/4 + 1/5 + ... =
2(1 - 1/2 + 1/3 - 1/4 + 1/5 - ... ) - 1 =
2ln(2) - 1

[@mr.doritos3669]

Omg Its blackpenredpen again!
Entertaining as always!

[@GSHAPIROY]

I've got a video idea for you. I know you made a video where you do 100 integrals in one take, but maybe next time you should do one 100x-iterated integral.

[@zerospeed6412]

I think ln(4/e) would look more pleasing.

[@anshumanagrawal346]

Lol ikr

[@not_vinkami]

You mean ln(4e^-1)

[@Your_choise]

Eh I feel like the e should be kept out of the ln

[@zerospeed6412]

@@not_vinkami wouldn't look as pretty

[@zerospeed6412]

@@Your_choise but it's sometimes fun to pull it out of hiding 😉

[@holyshit922]

S(x) = sum(S_{n}x^n,n=0..infinity) = ((1+x)ln(1+x)-x)/(x^2(1-x))
If you extract nth coefficient from series expansion of S(x)
and then calculate limit(S_{n},n=infinity) you will get the answer
I have got S(x) by integration of geometric series and then by generating function for partial sum

[@captainhd9741]

I notice that there is a pattern with the denominators 2 6 12 20 so add 4 then add 6 then add 8 and this second series is just add 2 each time. Is this relevant to the word/definition “telescoping”?

[@soatnod]

You can try to watch mind your decisions' telescoping sum videos

[@samsam4455]

Dear blackpenredpen,
I would appreciate if you can check the convergence of the séries: ((ln n)^n)/n! And (n^ln n)/n!. Many thanks Sir

[@samsam4455]

@Oily Macaroni i did it with ratio test but it is not "elegant" enough

[@stapler942]

I wonder if mathematicians have a rigorous definition of "...and so on." For example, how many terms of a sum are necessary to deduce the generating formula? 😉

[@spaghetti1383]

1/2-1/6+1/12-...+(-1)^(n+1)/(n(n+1))+...
I don't know if this is rigorous but it is practical. If you really want to be rigorous, you just would use summation notation.

[@engelsteinberg593]

Infinite many terms.

[@SteelM4chine]

2nd

[@felixnotkjellberg]

Me who literally forgot how to do this type of question my teacher taught me

[@advaykumar9726]

Engineers would be like: Now by the fundamental theorem of engineering the greatest integer function of e is 2 so the answer is 2*1-1=1!

[@davidbrisbane7206]

I'll just pretend there is no right answer and then right down and answer and claim it is correct. Too simple.