Making Sense of Ramanujan's Infinite Sum for General Audience. 

No ProbleZ

An example would be that 1/3 = 0.333… that x 3 would be .9999… but 1/3 x 3 should be 3/3

but if you do the same think for 0,888...=1 its the same but we fond that 0,999...=1 so how can 0,888...=0,999...=1 ?

this wont work for 0.88888

@@adam-denis This won't work cuz I can say 0.9 exists between 0.88888 and 1 na

​​@@mridulacharya8250There exists infinite real number between two distinct real numbers..then how did you concluded there is no real number between 0.999999.... and 1 ??

Welcome aboard!

me 2 lol

Me too

What type of assumption should be?

First, one need to prove that the series converges to a number, but the series of natural numbers diverges that is infinite as it should be

Beacuse of support from people like you this channel is thriving

you can still join the discord server

Thank you

Infinity is not a number

​@@fauzanree1983that's what he is saying.. infinity isn't distributive or associative

@@fauzanree1983 I think Infinite is higher dimension of Numbers

A guy playing with math, breaking all well-agreed definitions and common sense. No, you can not add positive numbers and get anything negative. There are definitions for convergence. What you can do is use math in your own way and get absurd results. Ramanujan turns in his grave being mentioned in this.

@@Param_Hayaran Infinity can be handled as a group or set and yes it does have distributive and associative properties but ONLY as a group. These do not translate when infinity is used as a number in an equation

U should solve quadratic in order to get the value of n n²+n=-1/6 n²+n+1/6=0 Now by Sridharacharya formula n= (-1+- rootunder 1/3)/2

Absurd.

​@@kartikdd8678 😂 Might be an absurd solution but it makes a lil amount of sense

@@sytherplayz ∞^2+∞-1/6=0 even worse

Absolutely correct

The best explanation I have seen. Here are couple of things - take the 1 + 2 + 3 infinite series and add 1 to it. Is this allowed? If allowed will the answer be 11/12? What’s the sum of other infinite series - 1 + 4 + 9 + 16 etc.

what do you think?

its the last one of the numbers which are between 0 and 1..

its 9.999999/9.99999 its not inf.

​@@mathOgenius I am not math expert, this is only my view, please correct me if any mistake Actually it is not infinity in that sense but think in this way , 0.99 and 0.999 both have different level of accuracy so 9S is like one step ahead than S , so both are not comparable

yes thats what was the whole point of the video, the infinity not having a sense of comparison is causing these proofs.

Infinity is weird and you should not expect that infinity will behave like a normal finite number. Infinity is not a number it can be a number but using it like a number will give you weird results.

Thanks for clarification. So that means the equal sign in 1=0.99999.... to infinite is only valid when you add the term +0.0000......1 behind the expression right? Otherwise we should not say 0.9999.... = 1, rather stick to its conventional mathemathical expression: limit n to infinite for 0.9999n approaches 1.

this is called the zeno paradox, In our world we can travel from A to B that means there must be a unit of length which cant be halved.

@@mathOgenius Thanks for pointing me to the right direction. I've googled Zeno's paradox and found the answer which confirms my surmise...

yes you can in countless ways disprove this, here is another very basic proof we know.. if x- integer , y- integer then x+y is also an integer so this equation is not correct..

@@mathOgenius what is it then? is "-1/12" just any random number, some that ramanujan came up because he thought "why not"? Could it actually be any number like 1234.25? I don't understand the whole concept. My brain would tell me the answer would be infinity and that there's no single grain of doubt about that

@Wrongidea807, There is no error in your thinking. I am glad that there is at least one sane mind in the comment section and that's you. First of all, 1+2+3+... is NOT equal to -1/12 as you correctly pointed out. Secondly, for the most part this video is non-sense. "Adding a symbol infinity at the end" does not make any sense and it is not the reason why things work out. I don't know your mathematical background, but here is the rough idea. Take the series 1+1/2^2+1/3^2+1/4^2+.... You can show that this series is convergent. In fact, you can replace the power 2 by anything slightly bigger than 1. That is, 1+1/2^(1.1)+ 1/3^(1.1)+... is also convergent and so on. But 1+1/2+1/3+1/4+... diverges to infinity. Therefore, one can make a function f(s)=1+1/2^s+1/3^s+... What we are saying is that f(s) is defined for all s>1. Now comes the non-trivial part, turns out we can take s to be complex number where the real part is >1 and the series is still convergent. This defines a nice function on certain part of the complex plane. This is an extremely nice function. Such functions are called analytic. Turns out this function can be extended to the whole complex plane (almost) by what is called an analytic continuation. The good news is that there is exactly one way to do this analytic continuation. This continued function is defined even when s=-1. When you take s=-1, this function has a value -1/12. But the key thing to note is that we are not saying that this function and 1+1/2^s+1/3^s+... are the same everywhere. These two quantities are same only when real part of s>1. But, if you ignore this and put s=-1 in the series, you get 1+2+3+4+... and hence one often abuses this relationship to say that 1+2+3+...=-1/12. (Ramanujan, I think (I may be wrong here, need to check), did not use the analytic continuation technique.)

​@@Wrongidea807It's not a random number. It is a constant in regularization. So 1+2+3+4+..+∞ diverges the sum will blow up into infinity. But if you remove the infinity there will be a constant (-1/12) that will remain.

Thank you!!

The 1 that you are talking about is not exactly equal to 1...it is approaching 1 remember....so you are essentially calculating the limit when the base is very very little less than 1 or little more than 1, Powered to infinity.....cus if it is actually much more than 1 then it'll lead to infinity and in case it is less than 1 then 0....so 1^infinity is essentially indeterminate form

It depends on which dimension. Not in the Euclidean space but there are Geometric spaces that allows a point of infinity to exist.

Because the first equation had an infinite number of nines after the decimal point,while the second one had a finite number of nines(three),which are not recurring.

second example was for non reoccurring one, to show what happens when we remove infinity

I would need the location of the changed digit otherwise its uncertain.

Negative apples must b the rotting apples added in the beginning... 😅

you can see the numberphile proof for now..

@@mathOgenius oh thanks!