Taylor series of $ln(1+x)$ Let me introduce my question : 
all of us know that $\displaystyle \ln(1+x) = \sum_{n=1}^{\infty}\frac{(-1)^{n}x^{n}}{n}$ for all $-1 <x \le 1$. 
But how can we get this bounds. Actually we should to check that $R_{n}(x) \to 0$, where $\displaystyle R_{n}(x) = \ln(1+x) - \sum_{k=1}^{n-1}\frac{(-1)^{n}x^{n}}{n}$. But using Lagrange form of remainder I can't prove that for this points $R_{n}(x) \to 0$. 
Another try was with $\displaystyle \sum_{n=0}^{\infty}(-1)^nx^n = \frac{1}{1+x}$ and integrate it by parts, so we get the logarithm. But if we want to check analytic function or not. How can we check it by remainder? Maybe it's better to use Cauchy form of remainder ? 
 A: With the ratio test we have, for $x\ne0$,
$$
\frac{|(-1)^{n+1}x^{n+1}/(n+1)|}{|(-1)^nx^n/n|}=
\frac{n+1}{n}|x|
$$
which converges to $|x|$. So we have convergence for $|x|<1$ and non convergence for $|x|>1$.
At $x=1$ the series converges by Leibniz's test, at $x=-1$ the series diverges.
Since differentiation term by term is allowed inside the interval of convergence and the derivative term by term of the given series is
$$
\sum_{n\ge1}(-1)^nx^{n-1}=\frac{1}{1+x}
$$
we can easily conclude that the given series sums to $\ln(1+x)$.
A: You might just use the theory of power series. 
Let us compute the radius of convergence of this series. Write the series as $$\sum_{n =1}^\infty c_n x^n$$ where $$c_n = \frac{(-1)^n}{n}$$
Then, we have that the radius of convergence $R$ is 
$$
R = \frac{1}{\limsup_{n\rightarrow\infty}\sqrt[n]{|c_n|}} = \frac{1}{\limsup_{n\rightarrow\infty}\sqrt[n]{\frac{1}{n}}} = \frac{1}{\lim_{n\rightarrow \infty} \frac{1}{\sqrt[n]{n}}} = 1 
$$
Thus, the series converges (pointwise) in (-1,1). 
