To solve this problem, we’ll analyze $f(x) = \frac{e^x - 1}{x}$ for $x \neq 0$ and $f(0) = 1$ using the Taylor series expansion of $f$ around $x = 0$.

Step 1: Taylor Series for $f(x)$

1. Start by expressing $e^x$ as a Taylor series centered at $x = 0$: $e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$

2. Then, $e^x - 1$ is: $e^x - 1 = x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$

3. Dividing $e^x - 1$ by $x$, we get: $f(x) = \frac{e^x - 1}{x} = 1 + \frac{x}{2!} + \frac{x^2}{3!} + \cdots = \sum_{k=0}^{\infty} \frac{x^k}{(k+1)!}$

Step 2: Taylor Series for $f'(x)$

To find $f'(x)$, differentiate $f(x) = \sum_{k=0}^{\infty} \frac{x^k}{(k+1)!}$ term by term: $f'(x) = \sum_{k=1}^{\infty} \frac{k x^{k-1}}{(k+1)!}$

Then, substitute $x = 10$ into $f'(x)$: $f'(10) = \sum_{k=1}^{\infty} \frac{k \cdot 10^{k-1}}{(k+1)!}$

Step 3: Evaluating the Series

The given expression to evaluate is: $\sum_{k=1}^{\infty} \frac{k \cdot 10^{k-1}}{(k+1)!}$ This series is exactly the form of $f'(10)$. Therefore, $\sum_{k=1}^{\infty} \frac{k \cdot 10^{k-1}}{(k+1)!} = f'(10)$

To approximate $f'(10)$, we can compute a few initial terms in the series for $f'(10)$, or evaluate it with more terms if precision is required.

Would you like me to compute the value of this sum up to a certain number of terms or explain further?

Here are some related questions to expand your understanding:

1. What would the Taylor expansion look like for other values of $x$, such as $x = -10$?
2. How does the convergence rate of this Taylor series vary with $x$?
3. How would we evaluate $f''(x)$ for a more detailed approximation?
4. Could we use numerical methods to approximate $f'(10)$ more quickly?
5. How does the convergence behavior change if we use different functions for $f(x)$?

Tip: When working with Taylor series, always consider how many terms are necessary to achieve a desired accuracy.