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
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1 Answers
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_summationUpvote·0· Reply ·2013-10-12 00:30:57
- Shaswata Roy I never said that it cannot be assigned a sum.Writing it in the form of a recursive relation is wrong.