IIT JEE , AIEEE Forum - Mathematics , Algebra - again an open challenge to all community members and experts

rudra.panda

And @Feynman sir please show some mercy to your juniors or to the mere beginners. I have posted the link in the next post. Kindly see it.

spideyunlimited

This sucks, man. So guys, never run after money, the more money you make, the closer you're tending to -1/12 lol

RyuAmakusa

sir we all know that if through some procedure we prove the fundamentals of mathematics is wrong then there is a mistake for sure in our procedure.i think i know what is wrong but it is really difficult to expresses it through a post,any way i'll give my best shot

Hence S(1+r) = r - r2 + r3 -....
or S(1+r) = r/(r+1) ---------------(1)
sir during the above processes we make an assumption that |r|<1
now in the next step S = r/(r+1)^2 now this function S is a cont. fn in it's domain and it's domain is (-1,1) so we cannot take the
lim r->1 we can only take lim[ r->1-](LHL) and it is 1/4 as u have shown now u have written
lim [r->1] S = lim[r->1] r - 2r^2 + 3r^3 - 4r^4+...
this function which u have sub in place of S is diff from the function S in step (1) both have diff domains.
we can't equate the lim[r->1] for both the fn. but we can equate the LHL at 1 for both the fn.
while equating the LHL's
lim[r->1-] r/(r+1)^ 2 = lim[r->1-] r - 2r^2 + 3r^3 - 4r^4+...
this is true but....but......
lim[r->1-] r - 2r^2 + 3r^3 - 4r^4+... NOT = 1 - 2 + 3 - 4 + ....
the LHL of the original fn S is not 1 - 2 + 3 - 4 + .... if we want to find the LHL then mult. and divide by (r+1) and u get the same result.
the problem was with that beautiful function
S = r - 2r2 + 3r3 - 4r4+... for this function LHL at 1 NOT= S(1)
it looks like a polynomial but it is NOT CONTINOUS

in other simple words the function S is written like this
_
| r/(r+1)^2 if -1S -
| r - 2r^2 + 3r^3 - 4r^4+... if r>=1 (or) r<= -1
_
if we rewrite the function in this way it clears the confusion.
i think this is the thing........sir is it correct????

RyuAmakusa

i think at the end it is not clear the rewrited function is

S= r/(r+1)^2 if -1<r<1 else r - 2r^2 + 3r^3 - 4r^4+...

pardesi

ok this is a quite a fundamental mistake that most peopel whos atrt with the riemann zeta functions end up doing!!

the riemann zeta function is defined by

$\zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^{s}}$

for Re(s) > 1 so u can't use the formuale relating riemann zeta functions to bernoulli numbers to find the value!!

also the sum clearly diverges because given an M>0 i can always find a n such that

$S_{n}=\sum_{i=1}^{n}i$ >M

so the sum clearly does not converge at any cost!!!

@Rudra the very fact that u assumed

$1+2+3+...=T$ . is wrong since the sum itself diverges as i proved above u can't assigna ny real number to it so that step is wrong!!!

it'a a common mistake so nothing to worry !!!

it's nice that u guys are doing such things but u shud know what u r doing!!

hmm it wud be very helpful if u go thru convergent and divergent series that is real analysiss atleast then talk about zeta functiosn and all otw taht may lead to misconceptions which are dangerous

GoNik

i think ryu has solved the prob......thanx..

GoNik

the sum of positive nos. cannot b negative....also the graph shows contradictory result.......hence it cannot b -1/12......

nknikhilesh1

How the sum can be negative.not understanding................

GoNik

and wat abt the graph???....not explained.....pleeeeeeez Experts take a look at this...

GoNik

the hint lies here i think:
Summation methods usually concentrate on the sequence of partial sums of the series. While this sequence does not converge, we may often find that when we take an average of larger and larger initial terms of the sequence, the average converges, and we can use this average instead of a limit to evaluate the sum of the series.

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