How can we find the value of $(1+x)(1-x+x^2-x^3+x^4....... \textrm{infinity})$. I think it will be 1 but not too sure of it. I think all the terms will get cancelled only 1 remains but how to show it?
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How can we find the value of $(1+x)(1-x+x^2-x^3+x^4- \dots)$ ?
This expression doesn't have a "value" per se. But this notation is usually interpreted as $\lim_{n \to \infty} (1+x)(1-x+x^2- \dots + (-x)^n)$ and value of this limit, whenever it exists, is considered to be a value of your expression (otherwise we say that expression doesn't have value at all).
I think it will be 1 but not too sure of it. I think all the terms will get cancelled only 1 remains but how to show it?
For the interpretation above, we get $\lim_{n \to \infty}(1+x)(1-x+x^2- \dots + (-x)^n) = \lim_{n \to \infty}(1-(-x)^{n+1})$. This limit exists (in traditional sense) iff $x \in [-1,1)$ and equals $0$ for $x=-1$, $1$ for other values of $x$. Note how "all the terms" never get cancelled.
Also, one can say that for $x < -1$ the limit equals $-\infty$, this statement has a number of possible interpretations. Be careful, though, since using $\pm \infty$ as a "value" can be tricky. Personally, I prefer to say that the limit simply doesn't exist in this case.
Abstraction
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I thought by simple multipllication we will get my result..... – Suprabha Nov 15 '16 at 09:11
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You can multiply polynomials, such as $1+x$ or $1-x+x^2-x^3$. But $1-x+x^2-x^3+\dots$ is not a polynomial and you should be careful around that $\dots$ part (for example, $2+4+8+16+\dots = 2*(1 + (2+4+8+16+\dots))$ but interpreting that as $z = 2(1+z)$ yields weird result, doesn't it?) A common ("first level") way to work with infinite sums is to take first $n$ terms and then look at the limit as $n$ increases. – Abstraction Nov 15 '16 at 09:25