To write the Taylor series for $f(x) = \frac{1}{x}$ centered at $a = 1$, we first need to find the derivatives of $f(x)$ and evaluate them at $x = 1$.

The Taylor series for a function $f(x)$ centered at $a$ is given by: $f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots$

For $f(x) = \frac{1}{x}$:

1. $f(x) = \frac{1}{x}$
2. $f'(x) = -\frac{1}{x^2}$
3. $f''(x) = \frac{2}{x^3}$
4. $f'''(x) = -\frac{6}{x^4}$
5. $f^{(4)}(x) = \frac{24}{x^5}$
6. And so on...

Evaluating these derivatives at $x = 1$:

1. $f(1) = 1$
2. $f'(1) = -1$
3. $f''(1) = 2$
4. $f'''(1) = -6$
5. $f^{(4)}(1) = 24$

Now, we can write the Taylor series centered at $a = 1$:

$f(x) = 1 - (x-1) + \frac{2}{2!}(x-1)^2 - \frac{6}{3!}(x-1)^3 + \frac{24}{4!}(x-1)^4 - \cdots$

Simplifying the coefficients:

$f(x) = 1 - (x-1) + (x-1)^2 - (x-1)^3 + (x-1)^4 - \cdots$

Thus, the Taylor series for $f(x) = \frac{1}{x}$ centered at $a = 1$ is:

$f(x) = \sum_{n=0}^{\infty} (-1)^n (x-1)^n$

This series converges for $|x-1| < 1$.

Do you want more details or have any questions?

Here are 5 more questions you might find interesting:

1. What is the radius of convergence for the Taylor series of $e^x$ centered at $x = 0$?
2. How do you find the Maclaurin series for $\sin(x)$?
3. Can you derive the Taylor series for $\ln(x)$ centered at $x = 1$?
4. What is the Taylor series for $\cos(x)$ centered at $x = \pi/2$?
5. How do you determine the interval of convergence for a given power series?

Tip: When calculating derivatives for Taylor series, it can be helpful to look for patterns in the derivatives to simplify the process.