$\dfrac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots$ this is valid for $x$ between $-1$ and $1$ not including the endpoints.
How can one show that the Lagrange remainder goes to zero as $n$ goes to infinity? Just trying to understand. Presumably it should go to zero.
On
Since $\frac{1-x^{n+1}}{1-x} =\sum_{k=0}^{n} x^k $, $\frac{1}{1-x} -\sum_{k=0}^{n} x^k =\frac{x^{n+1}}{1-x} $.
What is needed is to show that, for any $x \in(-1, 1)$, $\frac{x^{n+1}}{1-x} \to 0 $ as $n \to \infty$.
For a fixed $x$, this is equivalent to showing that $x^n \to 0$.
It is enough to show this for $x > 0$ (why?).
Watch closely - the fingers never leave the hands.
Let $x = \frac{1}{1+b}$ where $b = \frac1{x}-1 > 0$. $(1+b)^n > nb$ (by Bernoulli's inequality) so $x^n =\frac{1}{(1+b)^n} < \frac1{nb} \to 0 $ as $n \to \infty$.
You don't need the "Lagrange remainder". Simply take the difference of $\frac{1}{1-x}$ and the partial sum $1+x+x^2+...+x^n$ and you will see that it goes to zero as $n$ goes to infinity.