While finding the Taylor Series of a function, when are you allowed to substitute? And why?
For example:
Around $x=0$ for $e^{2x}$ I apparently am allowed to substitute $u=2x$ and then use the known series for $e^u$. But for $e^{x+1}$ I am not allowed to substitute $u=x+1$.
I know the technique for finding the Taylor Series of $e^{x+1}$ around $x=0$ by taking $e^{x+1}=e\times e^x$. However, I am looking for understanding and intuition for when and why it is allowed to apply substitution.
Note: there are several question that are similar to this one, but I have found none that actually answers the question "why"; or that shows a complete proof.
EDIT: Thanks to the answer of Markus Scheuer I should refine the question to cases where the series is finite, for example $n\to3$
Best Answer
We know $f(u)=e^u$ can be represented as Taylor series convergent for all $u\in\mathbb{R}$, i.e. the radius of convergence $R=\infty$. \begin{align*} f(u)=e^u=\sum_{n=0}^\infty \frac{u^n}{n!}\qquad\qquad\qquad u\in \mathbb{R} \end{align*}
We obtain \begin{align*} f(2x)=e^{2x}=\sum_{n=0}^\infty \frac{(2x)^n}{n!}\qquad\qquad\qquad x\in \mathbb{R} \end{align*}
We obtain \begin{align*} f(x+1)=e^{x+1}=\sum_{n=0}^\infty \frac{(x+1)^n}{n!}\qquad\qquad\qquad x\in \mathbb{R} \end{align*}
Conclusion: We can use any substitution for convenience as long as we respect the radius of convergence.