1. ## Power Series Expansion

Find the power series expansion of $ln(1+x)$ and hence derive the formula discovered in 1705 by Wilhelm Leibniz which states that

$ln2 = 1-1/2+1/3-1/4+1/5-1/6+...$

~Much appreciate the help

2. $y = \ln(1+x)$

$y' = \frac{1}{1+x}$

$y' = \frac{1}{1+x} = \frac{1}{1 - (-x)} = 1 - x + x^2 - x^3 + x^4 - x^5 + ...
$

integrate term for term ...

$y = C + x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ...$

at $x = 0$ , $\ln(1+x) = 0$ ... $C = 0$

$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ...$

$\ln(1+1) = \ln(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ...$

3. Originally Posted by Wildharmony
Find the power series expansion of $ln(1+x)$ and hence derive the formula discovered in 1705 by Wilhelm Leibniz which states that

$ln2 = 1-1/2+1/3-1/4+1/5-1/6+...$

~Much appreciate the help
Let $S_n=a+ar+ar^2+\cdots+ar^{n-1}$, so $rS_n=ar+ar^2+ar^3+\cdots+ar^n$

So $S_n-rS_n=a-ar^n\implies{S_n=\frac{a-ar^n}{1-r}}$

Now $S_n=a+ar+ar^2+\cdots+ar^{n-1}=\sum_{n=0}^{n-1}ar^n$

So now consider $\lim_{n\to\infty}S_n=\sum_{n=0}^{\infty}ar^n=\lim_ {n\to\infty}\frac{1-ar^n}{1-r}$

Now consdier that if $1\leqslant|r|$ this limit diverges to infinity, but if $|r|<1$ then $\lim_{n\to\infty}S_n=\frac{a}{1-r}$

$\therefore\forall{r}\backepsilon|r|<1~\sum_{n=0}^{ \infty}r^n=\frac{1}{1-r}$

Now consider that $|r|<1\implies|-r|<1$

So $\forall{r}\backepsilon|r|<1~\frac{1}{1-(-r)}=\frac{1}{1+r}=\sum_{n=0}^{\infty}(-r)^n=\sum_{n=0}^{\infty}(-1)^nr^n$

So now consider

\begin{aligned}\int_0^1\frac{dx}{1+x}&=\lim_{\varp hi\to{1}^-}\int_0^{\varphi}\frac{dx}{1+x}\\
&=\lim_{\varphi\to{1^-}}\int_0^{\varphi}\sum_{n=0}^{\infty}(-1)^nx^ndx\end{aligned}

Now we have shown that the interval of convergence of $S_{\infty}$ is $(-1,1)$, thus on thatinterval it is uniformly convergent, thus on $(0,\varphi)\subset(-1,1)$ the integral and summation be switched.

So \begin{aligned}\lim_{\varphi\to{1^-}}\int_0^{\varphi}\sum_{n=0}^{\infty}(-1)^nx^ndx&=\lim_{\varphi\to{1^-}}\sum_{n=0}^{\infty}\int_0^{\varphi}(-1)^nx^ndx\\
&=\lim_{\varphi\to{1^-}}\sum_{n=0}^{\infty}\frac{(-1)^n\varphi^{n+1}}{n+1}\\
&=\lim_{\varphi\to{1^-}}\sum_{n=1}^{\infty}\frac{(-1)^n\varphi^{n}}{n}\end{aligned}

Now notice that this series interval of convergence is $(-1,1]$, so by abel's theorem

$\lim_{\varphi\to{1^-}}\sum_{n=0}^{\infty}\frac{(-1)^n\varphi^n}{n}=\sum_{n=0}^{\infty}\frac{(-1)^n}{n}$

\begin{aligned}\therefore\int_0^1\frac{dx}{1+x}&=\ ln(1+x)\bigg|_{x=0}^{x=1}\\
&=\ln(2)\\