Show that ln(1+x)=x- dfrac x²2+ dfrac

Show that $\ln(1+x)=x-\dfrac{x^2}2+\dfrac{x^3}3-\cdots+(-1)^{n-1}\dfrac{x^n}n+R_n(x)$, and find an expression for the remainder $R_n(x)$.

Step 1: Derivatives of $f(x)=\ln(1+x)$

Directly: $f'(x)=(1+x)^{-1}$, $f''(x)=-(1+x)^{-2}$, $f'''(x)=2(1+x)^{-3}$, $f^{(4)}(x)=-6(1+x)^{-4}$; by induction, for $k\ge1$: \[f^{(k)}(x) = (-1)^{k-1}(k-1)!\,(1+x)^{-k}. \tag{1}\]

Step 2: Evaluate at $x=0$

\[ f^{(k)}(0) = (-1)^{k-1}(k-1)!, \qquad k\ge1, \qquad f(0)=0. \tag{2} \]

Step 3: Maclaurin formula

Substituting (2) into $f(x)=\sum_{k=1}^n\dfrac{f^{(k)}(0)}{k!}x^k+R_n(x)$, and simplifying $(k-1)!/k!=1/k$: \[\ln(1+x) = x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\cdots+(-1)^{n-1}\frac{x^n}n+R_n(x). \tag{3}\]

Step 4: Lagrange form of the remainder

With $\xi=\theta x$, $\theta\in(0,1)$, and using (1) at $k=n+1$: \[R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!}x^{n+1} = \frac{(-1)^n}{n+1}\cdot\frac{x^{n+1}}{(1+\theta x)^{n+1}}. \tag{4}\]

Step 5: Convergence for $x\in(-1,1]$

For $|x|\le1$, $x\neq-1$, one has $1+\theta x>0$ and $\left|\dfrac x{1+\theta x}\right|\le1$ (checked separately on $[0,1]$ and on $(-1,0)$, using $\theta\lt 1$), so

\[ |R_n(x)| \le \frac1{n+1} \to 0 \quad (n\to\infty). \tag{5} \]

Hence the series converges to $\ln(1+x)$ for $x\in(-1,1]$.

Step 6: Verification

\[x=0.5,\ n=3\text{: exact remainder }\ln1.5-(x-\tfrac{x^2}2+\tfrac{x^3}3)\approx-0.01120,\text{ and }|R_3|\le\tfrac14=0.25\ \checkmark\]