In the MacLaurin series of $\ln(1+x)$, why must $x$ lie in $(-1,1]$? On computing the derivatives and substituting $x=0$, I do not see why there is any problem with $x>1$.
Of course I understand that $x>-1$, since the domain of the logarithm is the set of positive real numbers.
So why doesn't the expansion work, for example, when $x=5$?
 A: The function $x\mapsto\ln(1+x)$ and all of its derivatives are well defined on $(-1,+\infty)$ ...
but the series $\sum_{n\ge1}\frac{(-1)^{n-1}}{n}x^n$ converges iff $x\in(-1,1]$ :


*

*For $\vert x\vert < 1$, you can apply the ratio test to the series $\sum_{n\ge1}\frac{\vert x\vert^n}{n}$ and see that the original series is absolutely convergent.

*For $x=1$, it is an alternate series whose term decreases (in modulus) to $0$

*For $\vert x\vert>1$, the sequence $\left(\frac{(-1)^{n-1}x^n}{n}\right)_{n\ge1}$ does not converge to $0$.
A: In this answer I'll explain that while the series diverges for $x > 1$ it still represents the function $\log(1+x)$ and one can actually use the divergent series to accurately evaluate $\log(1+x)$. In case of standard functions there are of course other methods to do this, e.g. to calculate $\log(6)$ you can evaluate $-\log\left(\frac{1}{6}\right)$ instead, the series converges for $x = -\frac{5}{6}$. But suppose that the series is all we have access to, then how do we reconstruct the function it represents in the region where the series diverges?
The reason why the radius of convergence is equal to 1 is the presence of the singularity at $x = -1$. A simple way to fix the problem is then to apply a transform to $x$ that moves that point farther away from the origin. If we define $z = \frac{x}{1+x}$ then the point $x = -1$ is mapped to infinity, and since $x = \frac{z}{1-z}$, you can substitute that for $x$ in the series expansion and expand what you get in powers of $z$ without problems:
$$\log(1+x) = z + \frac{z^2}{2} + \frac{z^3}{3} + \frac{z^4}{4}+\cdots$$
Since $x = 5$ corresponds to $z = \frac{5}{6}$, the series converges. In fact it is precisely the series we would have gotten had we evaluated $-\log\left(\frac{1}{6}\right)$ by putting $x = -\frac{5}{6}$ and multiplied the result by minus 1. But what we've done above is fundamentally different as we've actually used the series in the divergent region, so the divergent series does actually contain the information about the value of the function.
The series we've obtained does now convergence but it doesn't converge all that fast, the radius of convergence is 1 an $z = \frac{5}{6}$ is close to $1$. So, the idea of mapping the singularity at $x = -1$ to infinity didn't yield a very good result, and that's due to the presence of another singularity at infinity. The mapping $z = \frac{x}{1+x}$ maps that singularity at infinity to $z = 1$.  We can then try to improve the convergence by using a different mapping that puts both singularities as far away from the point where we want to evaluate the function as is possible. 
In this case, if we use the transform:
$$x = \frac{2 z}{3 - z}$$
then that puts the singularities at $z = \pm 3$, then we get the series:
$$\log(1+x) = \frac{2 z}{3}+\frac{2 z^3}{81}+\frac{2 z^5}{1215}+\frac{2 z^7}{15309}+\frac{2 z^9}{177147}+\cdots$$
and $x = 5$ corresponds to $z = \frac{15}{7}$, the above series converges a lot faster than the previous series for $\log(6)$.
A: A valid Taylor expansion also needs the reminder to converge to zero.
Pls check the precise definition of Taylor Therom:
Taylor's theorem： Let k ≥ 1 be an integer and let the function f : R → R be k times differentiable at the point a ∈ R. Then there exists a function hk : R → R such that
${\displaystyle f(x)=f(a)+f'(a)(x-a)+{\frac {f''(a)}{2!}}(x-a)^{2}+\cdots +{\frac {f^{(k)}(a)}{k!}}(x-a)^{k}+h_{k}(x)(x-a)^{k},}\\
{\displaystyle {\mbox{and}}\quad \lim _{x\to a}h_{k}(x)=0}$
https://en.wikipedia.org/wiki/Taylor%27s_theorem#:~:text=In%20calculus%2C%20Taylor's%20theorem%20gives,Taylor%20series%20of%20the%20function.
