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Math Help - Power Series Expansion

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    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
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  2. #2
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    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 + ...<br />

    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} + ...
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Wildharmony View Post
    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}\\<br />
&=\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\\<br />
&=\lim_{\varphi\to{1^-}}\sum_{n=0}^{\infty}\frac{(-1)^n\varphi^{n+1}}{n+1}\\<br />
&=\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}\\<br />
&=\ln(2)\\<br />
&=\sum_{n=0}^{\infty}\frac{(-1)^n}{n}\quad\blacksquare\end{aligned}
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