A nice infinite series courtesy of Ramanujan

[@SuperSilver316]

In my readings on the digamma function, series where the terms are rational functions can expressed as a finite sum of digamma functions evaluated at wherever the series is undefined multiplied by the appropriate constants, it’s actually such a cool formula!!

[@Risu0chan]

From the fraction decomposition, I solved it differently. Since each term is a harmonic series H(n) in 1/n, it diverges. However we have a nice approximation γ + ln(n) + o(1). By manipulating each harmonic series, it's possible to have a nice sum of H(2n) and H(4n). Finally all γ and all ln(n) cancel out. At the limit, the o(1) goes to zero, and what remains is the correct solution -½+¾ln2.

[@yoav613]

Nice! Ramanujan solved so many crazy and hard infinite sum and products,so i guess he solved this one in his head when he could'nt sleep 🤣💯

[@maths_505]

Indeed

[@grigoriefimovitchrasputin5442]

That's so cool to see people who loves maths and do that kind of content. Thank you so much for everything !

[@viniciusmoretti]

Let's call this sum inteegral.

[@Teja26051995]

These manipulations are cool, but how do we know they are justified? The infinite series for ln 2 is conditionally convergent. Referring to Tom Apostol's Mathematical Analysis, Section 8.6(Inserting and Removing Parentheses), the two series we've split into are parenthesizations of the ln 2 series. As such, their partial sums are subsequences of the sequence of partial sums for the ln 2 series, and hence converge to the same limit. To justify the splitting of the original sum, we probably need to note that the two parts converge, and their term-by-term addition also converges to the sum of limits. Just wanted to split some hairs to ensure we're rigorous here..

[@djbj1993]

How can you justify the step at 5:12 where you split the series in two? As far as I know the series needs to be absolutely convergent, which it obviously is not...

[@closenoor9263]

0:48 it's (2n-1)(2n+1)

[@zeggwaghismail827]

Very excellent. Thank you very much. Greetings from Morocco.

[@tani6990]

Common Ramanujan W

[@MrWael1970]

Thank you for your effort.

[@Abdouaep]

i love apples

[@bilkishchowdhury8318]

I discovered this formula: ln2 ÷4 = sum (1÷(n^3-n)) where n is the odd integers multiplied by 2

[@ranjithkumararunachalam3844]

Where I learn about series , calculas,....... As self study.

[@carlosgiovanardi8197]

Are there other sums similar to this one?

[@carlosgiovanardi8197]

I was playing with WA. And instead of 4, i used 5, 6, 7, 8, 9. Fun results were obtained.

[@santoshsubedi9471]

Please make a video on proof of theorem sent by Ramanujan to GH hardy in his letter

[@carlosgiovanardi8197]

next time the foxtrot series!!

[@maths_505]

Oh the one from the comics! Sure That can be evaluated nicely using residues

[@skakistisA]

Spoiler alert: This approximately 0.02

[@giuseppemalaguti435]

Ho fatto semplicemente con A/4n+B/4n-1+C/4n+1...e risulta( 1/2)(ln2-1)...ma perché é negativo? Boh, non capisco.. Forse in una serie a segni alterni(A, B, C hanno segni discordi), come prendi gli addendi, dà somme diverse...ho visto il video,e come sospettavo,tu hai preso gli addendi diversamente da me,facendo somme e sottrazioni qua e là..mah..??... I coefficienti A, B, C sono corretti, - 1,1/21/2...per cui risulta.. - 1/4-1/8-1/12-1/16....+1/6+1/14+1/22+1/30....+1/10+1/18+1/26+1/34...=-1/4+1/6-1/8+1/10-1/12+1/14-1/16+1/18-1/20....=(-1/2)(1/2-1/3+1/4-1/5+1/6-1/7+1/8-1/9...=(-1/2)(1-ln2)..e cosa mi dici?

[@primsiren1740]

Io non sono esperto però la somma no e convergente assolutamente. Puoi manipularla per prendere qualsiasi valore che vuoi. C'è bisogno di essere molto attento con somme come questo. Se metti "convergenza condizionale" in Google sicuramente puoi capire di più.

[@erfanmohagheghian707]

Ok Bruh. Don't lie to me that you do not delete comments which include links to other videos Lol. I'll send you screenshot if you're gonna insist on your lie. It's totally fine sir. It's your channel, your property. I wish you the best. I just thought that we could potentially collaborate. Have a good one.

[@maths_505]

Share the link right here as a reply to my comment. Lemme see how tf it gets deleted

[@erfanmohagheghian707]

@@maths_505 kzbin.info/www/bejne/j3OsdZ-Djs-XftE Worth to note that I learnt this method from Dr. Penn

[@maths_505]

Nd dude you pissed me off last time you pulled that deleted comment stunt. I just checked yt analytics. There's no comment from you in the published or held for review tabs. Nd go ask Daniel, the kid who makes incredible calculus videos. I tried to promote his videos myself by creating community posts linking to his videos. You're welcome to link comments back to your videos and even tag me in titles about coming up with better ways to solve the problem but don't post that slanderous BS that I'm deleting your comments. I don't know if it's a technical glitch and I dont care. Post the link NOW as a comment and if it doesn't work just tag me in the video title about how you came up with a better way to solve this problem. Got it kid.....

[@maths_505]

There you go that's the link....I don't see it deleted

[@erfanmohagheghian707]

@@maths_505 I believe you when you say there must be a technical problem. You're doing a great job and I have learnt things from you. I would love us to collaborate and I will of course put the link to your videos in my description. Honestly this is more fun and math love to me than trying to get views and I'm sure there are so many people out there who could do same stuff much better than I do.