Question Number 103219 by Dwaipayan Shikari last updated on 13/Jul/20

1+1+1+1+1+1+1+....=S_n   S_n =1+1+1+1+1+1+1+....  2S_n =    2 +      2   +     2+.......  .......... subtracting  −S_n =1−1+1−1+1−1+1−1+1−1+....  −S_n =(1/2)     S_n =−(1/2)    I have found this while experiment . I know the sum diverges  but is it pretty cool?   Kindly rectify me if there is any fault on this non rigorous  process  I have found some Ramanujan proof  S_n =1+2+3+4+5+6+7+...  4S_n =     4+   8   + 12+...       −3S_n =1−2+3−4+5−6+7−8+......  −3S_n =(1/4)  S_n =−(1/(12))    Ramanujan had done this on his notebook

Commented byJDamian last updated on 13/Jul/20

I always have been told   ∞−∞  is  undefined.  In this case S_n =∞ ⇒ 2S_n =∞ ⇒  S_n −2S_n = ∞−∞

Commented byDwaipayan Shikari last updated on 13/Jul/20

I know what do you want to say sir.But If we don′t take S_(n ) as  infinite then what will happen?

Commented byPRITHWISH SEN 2 last updated on 13/Jul/20

First of all ∞ is not a number so ∞−∞=0  or ∞+∞=2∞ is useless. ∞ is a conception it   is a symbol which represents the number which   is greater than any large number what you can imagine  ∵ it is not any arithmetic number that is why    no arithematic operations (such as +,−,÷,×)  can be done with ∞.  Now as far series is concerned the addition or  substraction of two series can be down only   when these two series converges to some finite  values.  Now there are large as well as smaller ∞  like for the set of integer  and for the set of real  numbers the ∞ of the former set(natural number set ) is smaller then the  ∞ of the later set (real number set ).I think this  much is enough for your doubt. Thank you.

Answered by prakash jain last updated on 13/Jul/20

ζ(0)=−(1/2),ζ(−1)=−(1/(12))  ζ(s) i think you already know.  I think you should read the links that  I sent earlier on this topic to get  some insight on why it works.  There a quite a bit of maths involved  and I really dont want to retype  material available.  I admire your curiosity and if you  put some more effort you will feel  more comfortable when looking  at sum of divergent series such  as 1+2+3+4+...=−(1/(12))

Commented byprakash jain last updated on 13/Jul/20

I will add a simple example of  gemometric series  f(x)=(1/(1−x))=1+x+x^2 +...   ∣x∣<1  using analytical continuity so  can sum  1+2+2^2 +2^3 +..=(1/(1−2))=−1  or you can write  S=1+2(1+2+2^2 +..)=1+2S  S=−1  I am not adding any mathematical  proofs here they are given in the  links that i shared before.

Commented byDwaipayan Shikari last updated on 13/Jul/20

I am always curious about these type of series.

Commented byDwaipayan Shikari last updated on 13/Jul/20

Sir  why two types of results are here     1−2+3−4+5−6+...=(1/4)  but logically it approches to −∞   or  1−2+3−4+....=−1−1−1−1−1−...=−(1+1+1+1+1+1+..)  =−(−(1/2))=(1/2)  i haven′t read the article fully  But it gives a overview that it is a arithmatic mean of some  results  I will read the full article  Thanking you sir for giving me this source.

Commented byDwaipayan Shikari last updated on 13/Jul/20

This is link which sir gave to me I want to share with everyone https://en.m.wikipedia.org/wiki/Analytic_continuation