Prove that the Taylor series converges to $\ln(1+x)$. Prove the following statement.

For $0 \leq x \leq 1$, the Taylor Series, $\displaystyle x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$ converges to $\ln(1+x)$

Any help will be greatly appreciated! 
Thank you!
 A: Alternatively, try integrating both sides of the algebraic identity (valid for $x\neq -1$):
$$\sum_{k=0}^{n-1} (-1)^kx^k=\frac{1}{1+x}+\frac{(-1)^{n-1}x^n}{1+x}$$
over a suitable interval and using elementary inequalities to bound the remainder term.
A: This is a personal proof of mine that I'd like you to try and fill.
Main Theorem:
$$\sum_{n =0}^\infty (-1)^{n} \frac{x^{n+1}}{n+1}=\log(1+x) \text{ ; } x \in(-1,1]$$
Let $n\in \Bbb N$ and $x \in \Bbb R$. Then the integral
T1. $$\alpha(n,x)=\int_0^x\frac{t^{2n+1}}{t+1}dt$$
satisfies the following relation:
$$ \alpha (n,x)-\alpha (n-1,x)  = \frac{x^{2n+1}}{2n+1} -\frac{x^{2n}}{2n}  $$ 
P. Note that
$$\frac {t^{2n+1}}{t+1}=t^{2n}-\frac{t^{2n}}{t+1}$$
Integrate, then integrate by parts.
T2.
$$\alpha \left( {n,x} \right) = \sum\limits_{v = 0}^{2n} {{{\left( { - 1} \right)}^v}
\frac{{{x^{v + 1}}}}{{v + 1}} - \log \left( {x + 1} \right)} $$
P. Use the formula in T1 in the following telescopic fashion:
$$\eqalign{
  & \alpha (n,x) - \alpha (n - 1,x) = \frac{{{x^{2n + 1}}}}{{2n + 1}} - \frac{{{x^{2n}}}}{{2n}}  \cr 
  & \alpha (n - 1,x) - \alpha (n - 2,x) = \frac{{{x^{2n - 1}}}}{{2n - 1}} - \frac{{{x^{2n}} - 2}}{{2n - 2}}  \cr 
  & \alpha (n,x) - \alpha (n - 1,x) + \alpha (n - 1,x) - \alpha (n - 2,x) = \frac{{{x^{2n + 1}}}}{{2n + 1}} - \frac{{{x^{2n}}}}{{2n}} + \frac{{{x^{2n - 1}}}}{{2n - 1}} - \frac{{{x^{2n}} - 2}}{{2n - 2}}  \cr 
  & \alpha (n,x) - \alpha (n - 2,x) = \frac{{{x^{2n + 1}}}}{{2n + 1}} - \frac{{{x^{2n}}}}{{2n}} + \frac{{{x^{2n - 1}}}}{{2n - 1}} - \frac{{{x^{2n-2}} }}{{2n - 2}} \cr} $$
reduce up to $\alpha(0,x)$ and evaluate this last (easy) integral.
T3.
$$\left| x \right| < 1 \Rightarrow \mathop {\lim }\limits_{n \to \infty } \alpha \left( {n,x} \right) = 0$$
P.
Since $\alpha >0 $ we can write
$$0 < \alpha \left( {n,x} \right) = \int\limits_0^x {\frac{{{t^{2n + 1}}}}{{t + 1}}dt}  $$
Note that $\frac{1}{x+1}$ is $1$ at $x=0$ and then decreases. So 
$$\frac{t^{2n+1}}{t+1}< t^{2n+1}$$
Now integrate from $0$ to $x$. Note the fact that $-1<x <1$ is essential for the limit to be $0$. The case $x=1$ is added by noting the series is the alternated harmonic series.
From the previous theorems we have that given any $\epsilon >0$, we can find a $\delta >0$ such that for any $n \geq n_0$
$$\eqalign{
  & \left| {\alpha \left( {n,x} \right)} \right| < \epsilon   \cr 
  & \left| {\sum\limits_{v = 0}^{2n} {{{\left( { - 1} \right)}^v}\frac{{{x^{v + 1}}}}{{v + 1}} - \log \left( {x + 1} \right)} } \right| < \epsilon  \cr} $$
A: Hint
Use the following theorem:
Theorem
Let $f(x)\in \textit{C}^{\infty}(-R,R)$. 
$$f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+...+\frac{f^{(k)}(0)}{k!}x^k+...$$
conveges in $(-R,R)$, if and only if  $\displaystyle{\lim_{n\to\infty}R_n(x)=0}$ for all $x\in(-R,R)$, where $R_n(x)$ is Lagrange's remainder:
$$R_n(x)=\frac{f^{(n+1)}{(\theta)}}{(n+1)!}x^{n+1}$$
