geometric series with negative exponent

Question

I wonder if it is possible to convert the infinite geometric series with negative exponent to a positive one? Is the calculation is correct ?

∑_(n=1)^∞▒x^(-n) =∑_(n=1)^∞▒〖(〖1/x)〗^n 〗=1/(1-1/x)=x/(x-1)

The pic is attached.

Answer 1

Yes, this is correct.

$1^\mathrm{st}$ step:

$$ x^{-n} = \left(\frac{1}{x}\right)^n $$ Here one can discuss whether $x\ne0$ must be, but this condition is contained in the next step anyway, which is why the discussion is also superfluous.

$2^\mathrm{nd}$ step:

If $|q|<1$:

$$ \sum_{n=1}^{\infty}{q^n}=\frac{1}{1-q} $$

Therefore, if $|\frac{1}{x}|<1 \Rightarrow |x|>1$

$$ \sum_{n=1}^{\infty}{\left(\frac{1}{x}\right)^n}=\frac{1}{1-\frac{1}{x}} $$

$3^\mathrm{rd}$ step:

$$ \frac{1}{1-\frac{1}{x}}=\frac{1}{\frac{x}{x}-\frac{1}{x}}=\frac{1}{\frac{x-1}{x}}=\frac{x}{x-1} $$

Put all steps together:

$$ |x|>1\quad\Rightarrow\quad\sum_{n=1}^{\infty}{x^{-n}}=\frac{x}{x-1} $$

Answer 2

The function $\frac{1}{1-x}$ has a pole at $x=1$, with

$$\lim_\limits{n\to 1^+} \frac{1}{1-x} = -\infty$$

$$\lim_\limits{n\to 1^-} \frac{1}{1-x} = \infty$$

For the reciprocal geometric series to work, we must have $|x|>1$.

The final result is valid, but requires $|x|<1$ to be expandable as a MacLaurin series.