Why is Taylor series expansion for 1 / ( 1 − x ) valid only for x ∈ ( − 1 , 1 ) ?

Question

Why is Taylor series expansion for   1      /    (  1  −  x  ) valid only for   x  ∈  (  −  1  ,  1  ) ?

Garagozzodgc · Accepted Answer

Well, we have that      s    n    =      ∑          k      =      0              n            x    k    =            1      −              x                  n          +          1                            1      −      x      Now, we look at   n  →  ∞      lim          n      →      ∞                  1      −              x                  n          +          1                            1      −      x        =      lim          n      →      ∞            ∑          k      =      0              n            x    k  Clearly, we&#039;re only concerned about       x          n      +      1      . If       |    x      |    &amp;gt;  1, then       x          n      +      1       grows large, and there is no limit. If       |    x      |    &amp;lt;  1, then   x  −  1  ≠  0 and       x          n      +      1       goes to 0 as   n  →  ∞, so in that case, we can assert that      lim          n      →      ∞                  1      −              x                  n          +          1                            1      −      x        =      1          1      −      x        =      ∑          k      =      0              ∞            x    k  If x=1, x−1=0 and we find ourselves in trouble. However, we can say that      ∑          k      =      0        n                      1        k              =  nin which case the sequence of partial sums has no limits. Finally, for x=−1, we have   (  −  1      )          n      +      1       which oscillates and has no limit.Thus,      1          1      −      x        =      ∑          k      =      0        ∞                      x        k            makes sense only for |x|&amp;lt;1.

Why is Taylor series expansion for 1/(1-x) valid only for x in (-1,1) ?

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