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Represent $1/(1-x)^2$ in a power series using the fact that $$\ln (x+1)= \sum_{n=1}^\infty (-1)^{n-1} \frac{x^n}{n}$$

First i derived $\ln(x+1)$ and its sum which equaled: $$1/(x+1)= \sum_{n=2}^\infty (-1)^{n-1} x^{n-1}$$ then I substituted $-x$ for $x$ giving me $$1/(1-x)= \sum_{n=1} x^n$$ then I derived this one more time to get the answer so I got $$1/(1-x)^2 = \sum_{n=2} nx^{n-1}$$ which is the wrong answer... The right answer is exactly the same but it starts from $1$. How?????? When we derive we add $+1$, so how it is wrong?

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There seems to be a misunderstanding in connection with the derivation of a series \begin{align*} A(x)=\sum_{n=0}^\infty a_nx^n \end{align*} We obtain \begin{align*} \frac{d}{dx}A(x)&=\sum_{\color{blue}{n=0}}^\infty na_nx^{n-1}=\sum_{\color{blue}{n=1}}^\infty na_nx^{n-1}\\ \end{align*}

Here we can increment the index $n$ by one, since the summand with $n=0$ is zero: $0a_0=0$ and has no contribution. Note, we do not add one to the index due to differentiation. It is simply due to the cancellation of the first summand in this case.

The derivation of the current series gives \begin{align*} \frac{d}{dx}\ln (x+1)&= \frac{d}{dx}\left(\sum_{\color{blue}{n=1}}^\infty (-1)^{n-1} \frac{x^n}{n}\right)\\ &=\sum_{n=1}^\infty n(-1)^{n-1} \frac{x^{n-1}}{n}\\ &=\sum_{n=1}^\infty (-x)^{n-1} \end{align*}

Since the index $n$ starts here with $n=1$, there is no summand with factor $n=0$ which can be cancelled.

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    soo when deriving a summation, we do not always just shift the index... we need to see what is inside the summation, correct? – 2017-02-15
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    sooo to be safe, if were not sure if the summand =0, we should just keep and index and not shift it, right?? – 2017-02-15
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    @exchangehelpforuni: Correct. – 2017-02-15
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    if n=0 has no contribution but we keep it, will our answer be considered wrong, or is it okay to jsut keep it?? – 2017-02-15
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    @exchangehelpforuni: Both variations are correct and quite ok. It is only a matter of convenience. – 2017-02-15
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You know that $$ \ln(1+x)=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^n}{n} $$ so $$ \ln(1-x)=-\sum_{n=1}^{\infty}\frac{x^n}{n} $$ Note the sum starts from $n=1$, not $2$.

Differentiating, $$ -\frac{1}{1-x}=-\sum_{n=1}^{\infty}x^{n-1} $$ that can be rewritten $$ \frac{1}{1-x}=\sum_{n=0}^{\infty}x^n $$ By differentiating again you get $$ \frac{1}{(1-x)^2}=\sum_{n=1}^{\infty}nx^{n-1}=\sum_{n=0}^{\infty}(n+1)x^n $$

Some explanations are apparently in order. Let's consider the general case $$ f(x)=\sum_{n=0}^{\infty}a_nx^n $$ Then the derivative can be written $$ f'(x)=\sum_{n=1}^{\infty}na_nx^{n-1} $$ because the constant term $a_0$ disappears and it's better to avoid writing the term $$ 0a_0x^{0-1} $$ which would be undefined at $0$.

In the case of $$ \ln(1-x)=-\sum_{n=1}^{\infty}\frac{x^n}{n} $$ we have $$ \ln(1-x)=\sum_{n=0}^{\infty}a_nx^n $$ where $$ a_n=\begin{cases} 0 & \text{if $n=0$}\\[4px] -\dfrac{1}{n} & \text{if $n>0$} \end{cases} $$ so the derivative is $$ -\frac{1}{1-x}=\sum_{n=1}^{\infty}na_nx^{n-1}= \sum_{n=1}^{\infty}(-1)x^{n-1}= -\sum_{n=0}^{\infty}x^n $$ the last equality by pulling the minus sign outside and changing indexing.

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    but when u diffrentiate, dont u add +1 to the index???? – 2017-02-14
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    when u derive, u add +1!!! – 2017-02-14
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    @exchangehelpforuni Why should I? What's the derivative of $x+x^2$? Constant terms disappear, non constant ones don't. – 2017-02-14
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    no i mean when u derive a series summation, u add 1 – 2017-02-14
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    for instance n=0, after u derive it will start at 1 – 2017-02-14
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    @exchangehelpforuni What you say is simply not true. You're going to make many mistakes by blindly follow (wrong) “rules”. – 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." index shifting when deriving,, isnt that true???? – 2017-02-14
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    @exchangehelpforuni I added some explanations. – 2017-02-14
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    i did not understand one single word – 2017-02-14
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    @exchangehelpforuni I will try to explain in other manner your problem: you have $-\frac1{1-x}=\frac{d}{dx}(-x-\frac{x^2}{2}-\frac{x^3}{3}-\ldots)=-1-x-x^2-\ldots=-\sum_{n=0}^{\infty}x^n$ – 2017-02-14