nuseldW4r

2022-12-03

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

Garagozzodgc

Expert

Well, we have that
${s}_{n}=\sum _{k=0}^{n}{x}^{k}=\frac{1-{x}^{n+1}}{1-x}$
Now, we look at $n\to \mathrm{\infty }$
$\underset{n\to \mathrm{\infty }}{lim}\frac{1-{x}^{n+1}}{1-x}=\underset{n\to \mathrm{\infty }}{lim}\sum _{k=0}^{n}{x}^{k}$
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 $x-1\ne 0$ and ${x}^{n+1}$ goes to 0 as $n\to \mathrm{\infty }$, so in that case, we can assert that
$\underset{n\to \mathrm{\infty }}{lim}\frac{1-{x}^{n+1}}{1-x}=\frac{1}{1-x}=\sum _{k=0}^{\mathrm{\infty }}{x}^{k}$
If x=1, x−1=0 and we find ourselves in trouble. However, we can say that
$\sum _{k=0}^{n}{1}^{k}=n$
in which case the sequence of partial sums has no limits. Finally, for x=−1, we have $\left(-1{\right)}^{n+1}$ which oscillates and has no limit.
Thus,
$\frac{1}{1-x}=\sum _{k=0}^{\mathrm{\infty }}{x}^{k}$
makes sense only for |x|<1.

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