Issue with Using Series to Approximate $\int_{1}^{e^2}\frac{e^x}{x}dx$

Question

I want to use a series to estimate $\int_{1}^{e^2}\frac{e^x}{x}dx$. When I write the series representation (a=0) in the integrand, divide by x term-by-term, integrate term-by-term, and put my answer back into a series form, I get a undefined first term (n=0 term). On the other hand, if I truncate to the third-order Taylor polynomial (a=0) first, then divide by x, then integrate term-by-term, I have no issues. Is there an order of operations issue here that I am not aware of? Why can't I truncate to $T_3(x)$ last?

Answer 1

So we agree that whether or not we truncate the Taylor series we have ($x > 0, \ n\geq 0$) $$\frac{e^x}{x} \approx \frac{1}{x} + \sum_{k=1}^n\frac{x^{k-1}}{k!}\hspace{2.5cm} \frac{e^x}{x} = \frac{1}{x} + \sum_{k=1}^\infty\frac{x^{k-1}}{k!}$$ Then : $$ \int_1^{e^2} \frac{e^x}{x} = \left[ \ln(x) + \sum_{k=1}^\infty \frac{x^{k}}{k!\cdot k}\right]_1^{e^2} = \overbrace{\ln(e^2)}^{=2} + \sum_{k=1}^\infty \frac{e^{2k}}{k!\cdot k} - \left(\overbrace{\ln(1)}^{=0} + \sum_{k=1}^\infty \frac{1}{k!\cdot k} \right) $$ And here I don't see where you have problem ?