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


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

$1 + 1 + 1 + 1 + 1 + 1 + 1 + . . . . = S_{n}$ $S_{n} = 1 + 1 + 1 + 1 + 1 + 1 + 1 + . . . .$ $2 S_{n} = 2 + 2 + 2 + . . . . . . .$ $. . . . . . . . . . s u b t r a c t i n g$ $- S_{n} = 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + . . . .$ $- S_{n} = \frac{1}{2}$ $S_{n} = - \frac{1}{2}$ $I h a v e f o u n d t h i s w h i l e e x p e r i m e n t . I k n o w t h e s u m d i v e r g e s$ $b u t i s i t p r e t t y c o o l ?$ $K i n d l y r e c t i f y m e i f t h e r e i s a n y f a u l t o n t h i s n o n r i g o r o u s$ $p r o c e s s$ $I h a v e f o u n d s o m e R a m a n u j a n p r o o f$ $S_{n} = 1 + 2 + 3 + 4 + 5 + 6 + 7 + . . .$ $4 S_{n} = 4 + 8 + 12 + . . .$ $- 3 S_{n} = 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + . . . . . .$ $- 3 S_{n} = \frac{1}{4}$ $S_{n} = - \frac{1}{12}$ $R a m a n u j a n h a d d o n e t h i s o n h i s n o t e b o o k$
Commented by JDamian last updated on 13/Jul/20

$I a l w a y s h a v e b e e n t o l d \infty - \infty i s$ $u n d e f i n e d .$ $I n t h i s c a s e S_{n} = \infty \Rightarrow 2 S_{n} = \infty \Rightarrow$ $S_{n} - 2 S_{n} = \infty - \infty$
Commented by Dwaipayan Shikari last updated on 13/Jul/20

$I k n o w w h a t d o y o u w a n t t o s a y s i r . B u t I f w e {d o n}^{'} t t a k e S_{n} a s$ $i n f i n i t e t h e n w h a t w i l l h a p p e n ?$
Commented by PRITHWISH SEN 2 last updated on 13/Jul/20

$First of all \infty is not a number so \infty - \infty = 0$ $or \infty + \infty = 2 \infty is useless . \infty is a conception it$ $is a symbol which represents the number which$ $is greater than any large number what you can imagine$ $∵ it is not any arithmetic number that is why$ $no arithematic operations (such as +, -, \div, \times)$ $can be done with \infty .$ $Now as far series is concerned the addition or$ $substraction of two series can be down only$ $when these two series converges to some finite$ $values .$ $Now there are large as well as smaller \infty$ $like for the set of integer and for the set of real$ $numbers the \infty of the former set (natural number set) is smaller then the$ $\infty of the later set (real number set) . I think this$ $much is enough for your doubt . Thank you .$
	Answered by prakash jain last updated on 13/Jul/20

	$ζ (0) = - \frac{1}{2}, ζ (- 1) = - \frac{1}{12}$ $ζ (s) i think you already know .$ $I think you should read the links that$ $I sent earlier on this topic to get$ $some insight on why it works .$ $There a quite a bit of maths involved$ $and I really dont want to retype$ $material available .$ $I admire your curiosity and if you$ $put some more effort you will feel$ $more comfortable when looking$ $at sum of divergent series such$ $as 1 + 2 + 3 + 4 + . . . = - \frac{1}{12}$
	Commented by prakash jain last updated on 13/Jul/20

	$I will add a simple example of$ $gemometric series$ $f (x) = \frac{1}{1 - x} = 1 + x + x^{2} + . . . ∣ x ∣< 1$ $using analytical continuity so$ $can sum$ $1 + 2 + 2^{2} + 2^{3} + . . = \frac{1}{1 - 2} = - 1$ $or you can write$ $S = 1 + 2 (1 + 2 + 2^{2} + . .) = 1 + 2 S$ $S = - 1$ $I am not adding any mathematical$ $proofs here they are given in the$ $links that i shared before .$
	Commented by Dwaipayan Shikari last updated on 13/Jul/20

	$I a m a l w a y s c u r i o u s a b o u t t h e s e t y p e o f s e r i e s .$
	Commented by Dwaipayan Shikari last updated on 13/Jul/20

	$Sir why two types of results are here$ $1 - 2 + 3 - 4 + 5 - 6 + . . . = \frac{1}{4}$ $but logically it approches to - \infty$ $or$ $1 - 2 + 3 - 4 + . . . . = - 1 - 1 - 1 - 1 - 1 - . . . = - (1 + 1 + 1 + 1 + 1 + 1 + . .)$ $= - (- \frac{1}{2}) = \frac{1}{2}$ $i {haven}^{'} t read the article fully$ $But it gives a overview that it is a arithmatic mean of some$ $results$ $I will read the full article$ $Thanking you sir for giving me this source .$
	Commented by Dwaipayan 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
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