<s>[INST] Binomial expansion of negative exponents. Let's say I have to expand $(1+x)^{-1}$ using binomial expansion.
Using the theorem, I get:
$$(1+x)^{-1} = 1-x+x^2-x^3+x^4-x^5+x^6+....+{\infty}$$
Substituting $x$ for $1$, I get:
$$\frac{1}{2}= 1-1+1-1+1-1+1+....+{\infty}$$
A similar result arises with higher power of the exponent
For $(1+x)^{-2}$ we get:
$$(1+x)^{-2} = 1-2x+3x^2-4x^3+5x^4-6x^5+7x^6+....+{\infty}$$
Substituting $x$ for $1$, I get:
$$\frac{1}{4}= 1-2+3-4+5-6+7+....+{\infty}$$
How does this makes sense? Help please!
 [/INST] The equality $$\sum_{n=0}^\infty x^n = \frac{1}{x+1}$$
is only true if $|x<1|$, so you cannot substitute $x=1$ into it and expect the result to work.
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