Taylor series of 1 over 1 minus x around x=10 and radius of convergence

Key takeaway: A Taylor series centered at a point other than 0 is easiest when you rewrite the function in terms of $(x-10)$. Then the usual geometric-series pattern appears with very little effort.

We want the Taylor series of

$f(x)=\frac{1}{1-x}$

centered at $x=10$.

Step 1: Rewrite using $(x-10)$

Let

$u=x-10.$

Then $x=10+u$, so

$1-x=1-(10+u)=-9-u=-(9+u).$

Thus

$\frac{1}{1-x}=-\frac{1}{9+u}=-\frac{1}{9}\cdot \frac{1}{1+u/9}.$

Step 2: Use the geometric series

Recall

$\frac{1}{1-w}=\sum_{n=0}^{\infty} w^n \quad \text{for } |w|<1.$

Here,

$\frac{1}{1+u/9}=\frac{1}{1-(-u/9)}=\sum_{n=0}^{\infty}\left(-\frac{u}{9}\right)^n,$

valid when $|u/9|<1$, i.e. $|u|<9$.

So

$\frac{1}{1-x}=-\frac{1}{9}\sum_{n=0}^{\infty}\left(-\frac{x-10}{9}\right)^n.$

Step 3: Write the series neatly

$\frac{1}{1-x}=\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{9^{n+1}}(x-10)^n.$

The first few terms are

$-\frac19+\frac{x-10}{81}-\frac{(x-10)^2}{729}+\cdots$

Step 4: Largest radius $R$

The singularity of $\frac{1}{1-x}$ is at $x=1$. The distance from the center $10$ to $1$ is

$|10-1|=9.$

So the largest radius is

$R=9.$

Therefore, the Taylor series converges for

$|x-10|<9.$

---

Pitfalls the pros know 👇 It’s easy to stop at $u=x-10$ and forget to convert back to $x$. Another trap is thinking the series should converge all the way up to $x=1$; the endpoint is excluded because the function itself blows up there.

What if the problem changes? If the center were $x=a$ instead, the same method would give a geometric series in $(x-a)$, and the radius of convergence would be the distance from $a$ to the singular point $x=1$, namely $|a-1|$.

Tags: Taylor series, radius of convergence, geometric series