represent the following in a power series

Question

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?

Answer 1

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 $$ \frac{1}{(1-x)^2}=\sum_{n=1}^{\infty}nx^{n-1}=\sum_{n=0}^{\infty}(n+1)x^n $$

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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.

Answer 2

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.