Expand 1/1+x² in ascending order of 1/x?

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Answer 1 · 2018-03-04T18:32:47

See below.

#1/(1+x^2) = (1/x^2)/(1/x^2+1)#

now for #abs x > 1#

#1/(1+1/x^2) = 1-1/x^2+1/x^4-1/x^6+ cdots + (-1)^n/(x^(2n))+ cdots#

then

#1/(1+x^2) =1/x^2 sum_(k=0)^oo (-1)^k/(x^(2k))#

which converges for #absx > 1#

NOTE

#1-a+a^2-a^3+ cdots+(-1)^n a^n = (a^(n+1)+1)/(a+1)#

then if #abs a < 1#

#lim_(n->oo) sum_(k=0)^n (-1)^k a^k = 1/(1+a)#