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When using the Maclaurin formula you get that $$\frac1{1-x} = \sum_{n=0}^\infty x^n$$ however, when you use the property that $$\ln(x+1)= \sum_{n=1}^\infty (-1)^{n-1} \frac{x^n}{n}$$ and then derive it getting: $$\frac{1}{x+1}=\sum_{n=1}^\infty (-1)^n x^n$$ then plugging in $-x$ for $x$, you get that $$\frac1{1-x} = \sum_{n=1}^\infty x^n$$ Why is the index wrong? When deriving you add $+1$

So why is it like this?

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    Possible duplicate of [represent the following in a power series](http://math.stackexchange.com/questions/2144214/represent-the-following-in-a-power-series) – 2017-02-14
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    @Maczinga but no one is answering there as it is a specific question... plus its diff before marking it as a "duplicate" read the question – 2017-02-14
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    It is pratically the same question... did you read my comment for the previous question? – 2017-02-14
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    $(x^n)'=nx^{n-1}$. – 2017-02-14

2 Answers 2

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You made a mistake in the manipulation of indices. $$\frac d{dx}\left(\sum_{n=1}^\infty(-1)^{n-1}\frac{x^{\color{red}{n}}}{n}\right)=\sum_{\color{blue}{n=1}}^\infty (-1)^{n-1}x^{\color{red}{n-1}}\stackrel{\color{blue}{k:=n-1}}=\sum_{\color{blue}{k=0}}^\infty (-1)^{\color{blue}{k}}x^\color{red}{k}=\sum_{k=0}^\infty (-x)^k$$

... or perhaps you are more familiar with the formula $\frac d{dx}\sum\limits_{n=h}^\infty a_nx^n=\sum\limits_{n=h-1}^\infty (n+1)a_{n+1}x^n$, from which $$\color{red}{\frac d{dx}}\left(\sum_{\color{red}{n=1}}^\infty(-1)^{\color{blue}{n-1}}\frac{x^n}{\color{blue}{n}}\right)=\sum_{\color{red}{k=0}}^\infty (-1)^{\color{blue}{k+1-1}}\frac{k+1}{\color{blue}{k+1}}x^{\color{red}{k}}=\sum_{k=0}^\infty(-x)^k\quad?$$

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    WAIT when u derived why didnt u add +1 to the index.. index shifting when deriving?? – 2017-02-14
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If $$\ln(1+x)=\sum_{n=1}^\infty (-1)^{n-1}\frac{x^n}{n}=x-\frac12x^2+\frac13x^3-\frac14x^4+\cdots$$ deriving, you get $$\frac{1}{1+x}=\sum_{n=1}^\infty(-1)^{n-1}x^{n-1}=1-x+x^2-x^3+\cdots$$ and plugging in $-x$, you get $$\frac{1}{1-x}=\sum_{n=1}^\infty x^{n-1}=1+x+x^2+x^3+\cdots$$

Your mistake is that you're changing the sum indices, note that $$\sum_{n=1}^\infty x^{n-1} = \sum_{n=0}^\infty x^n \quad \text{but} \quad \sum_{n=1}^\infty x^{n-1}\neq \sum_{n=1}^\infty x^{n}$$

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    WAIT when u derived in the 2nd line why didnt u add +1 to the index.. index shifting when deriving?? @flytothesurface – 2017-02-14
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    When you derive a power series, you just have to derive term by term. I wrote some terms to made you realise that you don't have to sum anything. – 2017-02-14
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    "Note that the start of the summation changed from n=0 to n=1, since the constant term a0 has 0 as its derivative." see when deriving, u need to change the index u do +1 – 2017-02-14
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    But you don't need to. It's better to look at the series you are deriving. By applying this kind of reasoning blindly, without looking at what you're deriving, can lead you to mistakes like that. – 2017-02-14
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    so when exactly do we add 1 to the index.. i was told by my instructor to add 1 whenever u derive the power series – 2017-02-14
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    If you add 1, you have to start the series at $n=0$. The series $\sum_{n=1}^\infty x^{n-1} $ and $\sum_{n=0}^\infty x^n$ are the same. – 2017-02-14
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    ln(x+1)=∑n=1 (−1)^(n−1) x^n/n if deriving it, do we then get.... 1/x+1=∑n=2 (−1)^(n-1) x^(n-1) ???????? – 2017-02-14