I've been taught that sum$(S)$ of first $n$ terms of G.P. is given by– $$S=\frac{a(1-r^n)}{1-r}$$ where, $a$ is first term of G.P., $r$ is common ratio and $r>1$ or $r<-1$
But, if we talk about infinite G.P., the sum $(S)$ is given by– $$S=\frac{a}{1-r}$$ where, $a$ is first term of G.P., $r$ is common ratio. But here, $|r|$$<$$1$.
And, the reason my teacher gave for the derivation of formula is that– as, $n$$\to$$\infty$, $r^n$$\to$$0$.
Doubt= Since $r^n$$≈$$0$ , So sum of infinite G.P. must also be an approximation.Isn't it????
The confusion might be with what limits are. If I ask you, "as $x$ goes to $\infty$, what happens to the value of $\frac{1}{x}+5$," what would you say? Well, the expression $\frac{1}{x} + 5$ approaches -- but never actually reaches -- $5$. You can verify this: there is no single $x$ for which $1/x = 0$.
In a similar way, the value of an infinite geometric series can be seen as a limit, namely, $$S = \frac{a}{1-r} = \lim_{n\to\infty} \sum_{i = 1}^n r^i.$$
There is no $n \in \mathbb{N}$ such that $\sum_{i=1}^n r^i = S$. But as we keep on increasing the number of terms in the sum, we get closer and closer to $S = \frac{a}{1-r}$.