# Whers d fallacy??

S=1-1+1-1+1-1....................... infinity

now

thus, S=1- (1-1+1-1+1-1....................... infinity)

thus S=1-S

thus 2S=1

S=1/2

2305
Shaswata Roy ·

Let,

\mathcal{S}=a+ar+ar^2+\cdots+ar^n=a+r\mathcal{S}-ar^{n+1}

If 0<|r|<1,

\lim_{n\rightarrow \infty}ar^{n+1}=0

Hence if n→∞,

\mathcal{S}=a+r\mathcal{S}

Therefore,this method can be applied iff the series is converging.The last term of the above series is either 1 or -1 and is not negligible.(i.e \lim_{n\rightarrow \infty}ar^{n+1}\neq0)

You can't write S=1-S

If we try to apply this trick on divergent or oscillating series(like the above one) we end up with weird results.As another example try applying this trick on a divergent series like,

\mathcal{S}=1+2+2^2+\cdots

You'll find that S is negative!

• Hari Shankar Under certain conditions divergent series can be assigned a sum: http://en.wikipedia.org/wiki/Ces%C3%A0ro_summation
• Shaswata Roy I never said that it cannot be assigned a sum.Writing it in the form of a recursive relation is wrong.