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
1−1+1−1+1−1+1−1+.....=(1/2) {But it diverges
1+1+1+1+1+1+1+......=−(1/2) {But it diverges
1+2+4+8+16+.....=−1 {But it diverges
1+2+3+4+5+6+7=−(1/(12)) {But it diverges
1−2+4−8+.....=(1/3) {But it diverges
1−2+3−4+5−6+.....=(1/4) {Is it a divergent?????
show that: cosθ + cos2θ + ....cos nθ= ((cos (1/2)(n +1)θ sin(1/2)nθ)/(sin (1/2)nθ))
Show that: sin θ + sin 2θ + ....+ sin nθ = ((sin (1/2)(n + 1)θ sin(1/2)nθ)/(sin (1/2)nθ))
where θ ∈ R and θ ≠2πk , k ∈Z
Sir MJS & Sir John Santu
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