# Using the Geometric Series to find where a series converges

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?

• How do you want us to answer if you don't tell us which function you are talking about? Concerning your final question, yes, it converges for all real numbers. – José Carlos Santos Aug 28 '19 at 21:03
• I put in the function – Burt Aug 28 '19 at 21:20

## 2 Answers

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 $$-2 \lt x \lt 2$$

• Since sin(x) and $e^x$ both converge for all real x, that means that the Taylor series is a good approximation for all real x, right? – Burt Aug 28 '19 at 21:18
• @burt - If $x$ is large than you will need a lot of terms for a good approximation. For example $\sin(10) \approx -0.544$ but you need to add up about sixteen terms in the Taylor series to get this to three decimal places. Meanwhile $\sin(100) \approx -0.506$ but you probably cannot approach this through the Taylor series without much more precision than usual – Henry Aug 28 '19 at 21:33
• I see. But technically it would converge for all x. – Burt Aug 28 '19 at 22:55
• And, if I use the Taylor series centered around 100 I'll get a much quicker approximation? – Burt Aug 28 '19 at 22:56
• @burt: The Taylor expansion of $\sin(x)$ about $x=100$ involves $\sin(100)$ and $\cos(100)$. If you want to find an approximation to $\sin(100)$ then it is perhaps better to do the expansion about $x=32\pi \approx 100.53$ – Henry Aug 28 '19 at 23:02

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 $$\mid x-1\mid\lt1$$.

• But how do you know when it converges? – Burt Aug 28 '19 at 23:16
• The geometric series has partial sum equal to $\dfrac {1-x^{n+1}}{1-x}$, which converges to $\dfrac 1{1-x}$, precisely when $\mid x\mid\lt 1$. Look at the numerator, and note that $x^{n+1}\to0$. – Chris Custer Aug 28 '19 at 23:21