Taylor theorem approximation

Question

$$e^x = \sum_{i=0}^n \frac{x^i}{i!} + \frac{x^{n+1}}{(n+1)!}e^{c_x}$$

is given based on Taylor's theorem. Where $$ c_x \in [0,x] $$

I understand that the statement below is true by Taylor's theorem approximation

$$e^x \approx \sum_{i=0}^n \frac{x^i}{i!} $$

However, I don't understand the statement below can be approximated based on the equation above.

$$\int_{0}^{1} \frac{e^x-1}{x} dx \approx \sum_{i=1}^n \frac{1}{i*i!} $$

How do you prove this approximation statement?

Answer 1

Hint.

Write the function $f$ as a power series and integrate term by term.

---

Notes.

It follows from $e^x\approx\sum_{k=0}^n\frac{x^k}{k!}$ that $$ \frac{e^x-1}{x}\approx\sum_{k=1}^n\frac{x^{k-1}}{k!} $$

Answer 2

\begin{align} \int_{0}^{1} \frac{e^x-1}{x} dx &\approx \int_{0}^{1} \frac{\sum_{i=0}^n \frac{x^i}{i!}-1}{x} dx \\ &= \int_{0}^{1} \frac{1+\sum_{i=1}^n \frac{x^i}{i!}-1}{x} dx \\ &= \int_{0}^{1} \sum_{i=1}^n \frac{x^{i-1}}{i!} dx \\ &= \sum_{i=1}^n \int_{0}^{1} \frac{x^{i-1}}{i!} dx \\ &= \sum_{i=1}^n \frac{1}{i\cdot i!} dx \\ \end{align}