<s>[INST] Taylor expansion of $\ln(1 + \frac{2^x}{n})$ I have a function $f = \ln(1 + \frac{2^x}{n})$, where $n \to \infty$ and $x \in (0, 1)$. 
I want to apply Taylor expansion at $a = 0$ to $f$.
I get $f = \dfrac{2^x}n-\dfrac{4^x}{2n^2}+o(\dfrac{4^x}{2n^2})$
My question is for what $x$ this approximation is true and why?
Wolfram Alpha says that it works for $|x| < 1$.
(I guess the answer for "why" is that the sum that we get in Taylor expansion is only converges for |x| < 1)
And I'm kinda confusing about that we expand it at $0$, then initially I thought it should only work for $x$ that are in $\epsilon$-neighborhood of $0$.
In general, I can expand the function at any point $a$, and this approximation is true $\forall x : \text{expansion converges}$?
 [/INST] Let $f(x) = \ln\left( 1 + \frac{2^{x}}{n}\right)$ then
\begin{align}
f'(x) &= \frac{\ln2 \, 2^{x}}{n + 2^{x}} \\
f''(x) &= \frac{n \, \ln^{2}2 \, 2^{x}}{(n+2^{x})^2} \\
f'''(x) &= \frac{n(n-1) \, \ln^{3}2 \, 2^{x} \, (n-2^{x})}{(n+2^{x})^3}
\end{align}
which leads to
$$f_{n}(x) = \ln\left(1 + \frac{1}{n}\right) + \left(\frac{x \, \ln2}{n+1}\right) + \frac{n}{2!} \, \left(\frac{x \, \ln2}{n+1}\right)^2 + \frac{n(n-1)}{3!} \, \left(\frac{x \, \ln2}{n+1}\right)^3 + \cdots$$
For large $n$ values this expansion can be seen as 
\begin{align}
f_{n}(x) &= \ln\left(1 + \frac{1}{n}\right) + \left(\frac{x \, \ln2}{n \, \left(1 + \frac{1}{n}\right)}\right) + \frac{1}{2! \, n} \, \left(\frac{x \, \ln2}{n \, \left(1 + \frac{1}{n}\right)}\right)^2 \\
& \hspace{10mm} + \frac{1-\frac{1}{n}}{3!} \, \left(\frac{x \, \ln2}{n \, \left(1+\frac{1}{n}\right)}\right)^3 + \cdots
\end{align}
This leads to
\begin{align}
\lim_{n \to \infty} f_{n}(x) = 0
\end{align}
</s>