Using the Geometric Series to find where a series converges

Question

I was given a function and told to use the geometric series to find the Taylor series. I was then asked where the series converges. What does this mean? How do I go about solving a problem like this?
The function I was given was:
$$\frac{1}{2-x}$$ However, in general, how should I solve such a problem?

Also, does the Taylor series only help for inputs near zero?
Is that what convergence means - that the Taylor Series works for those numbers? If yes, then does the sine function converge for all real numbers?

Answer 1

As examples which might help :

• $1+x+x^2+x^3+\cdots$ only converges for $-1 \lt x \lt 1$ and then (as a geometric series) is equal to $\frac1{1-x}$ and is its Taylor series around $x=0$

• $1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$ converges for all real $x$ and is equal to $e^x$ and is its Taylor series around $x=0$

Your case of $\sin(x) = \frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$ is related in many ways to the second example and also converges for all real $x$

Added: Your later addition of $\frac{1}{2-x}$ which you might expand to $\frac12 +\frac{x}{2^2} +\frac{x^2}{2^3} +\cdots$ is similar to the first example and only converges when $-2 \lt x \lt 2$

Answer 2

Let's use your example: $\dfrac1{2-x}=\dfrac 12\dfrac 1{(1-\frac x2)}=\dfrac 12\sum_{n\ge0}(\dfrac x2)^n$, and converges when $\mid\dfrac x2\mid\lt1$, or $\mid x\mid\lt2$. This is the Taylor series at $0$.

You can use the same sort of procedure to get the Taylor series at $1$, say: $\dfrac 1{2-x}=\dfrac 1{1-(x-1)}=\sum_{n\ge0}(x-1)^n$, which converges when $\mid x-1\mid\lt1$.