Why is Taylor series expansion for 1 / ( 1 − x ) valid only...

nuseldW4r

nuseldW4r

Answered

2022-12-03

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

Answer & Explanation

Garagozzodgc

Garagozzodgc

Expert

2022-12-04Added 9 answers

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're only concerned about x n + 1 . If | x | > 1, then x n + 1 grows large, and there is no limit. If | x | < 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 = n
in 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|<1.

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