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This question is continuing from the previous question here:

Power Series representation of $\frac{1+x}{(1-x)^2}$

I am trying to calculate the power series representation of the equation:

$$ \begin{align} f(x) = \frac{1+x}{(1-x)^2} \end{align} $$

My workout is as follow:

$$ \begin{align} \frac{1+x}{(1-x)^2} = \frac{1}{(1-x)^2} + \frac{x}{(1-x)^2} \end{align} $$

For $\frac{1}{(1-x)^2}$: \begin{align} \frac{1}{(1-x)^2} &= \frac{d}{dx} \frac{1}{1-x}\\ &= \frac{d}{dx} \sum_{n=0}^{\infty} x^n \\ &= \sum_{n=1}^{\infty} nx^{n-1} \\ &= \sum_{n=0}^{\infty} (n+1)x^n \end{align}

For$\frac{x}{(1-x)^2}$: $$ \begin{align} x \frac{1}{(1-x)^2} &= x \sum_{n=0}^{\infty}(n+1)x^n \\ &= \sum_{n=0}^{\infty} (n+1) x^{n+1} \end{align} $$

Therefore, $$ \begin{align}\frac{1+x}{(1-x)^2} = \sum_{n=0}^{\infty} (n+1)x^n+\sum_{n=0}^{\infty} (n+1) x^{n+1} = \sum_{n=0}^{\infty}(n+1)(x^n + x^{n+1}),\end{align} $$ where range of convergence is $x\in[-1,1)$. When $x=-1$, $(x^n + x^{n+1})$ becomes $0$, and $(\infty)(0) = 0$.

However, the model answer is $\sum_{n=0}^{\infty} (2n+1) x^n$, where range of convergence is $x\in (-1,1)$.

I do not understand what is wrong with my calculation. Any advice will be appreciated!

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    $\begingroup$ What is incomplete in your approach is that $\sum\limits_{n=0}^{\infty}(n+1)(x^n + x^{n+1})$ is not the canonical form of a power series, which should rather be expressed as $\sum\limits_{n=0}^{\infty}a_nx^n$ for some suitable sequence $(a_n)$. $\endgroup$
    – Did
    Feb 14, 2016 at 10:48
  • $\begingroup$ What should I do? In addition, term by term, the terms in my solution and the model answer are different. $\endgroup$
    – Joseph
    Feb 14, 2016 at 10:50
  • $\begingroup$ Was my solution wrong? The first term in my solution is $1+x$ but the first term in the model answer is $1$. $\endgroup$
    – Joseph
    Feb 14, 2016 at 10:52
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    $\begingroup$ Why does it seems that $f(x)\neq \sum_{n=0}^{\infty}(n+1)(x^n + x^{n+1})$? You conclude this by looking at the first term of the sum. In your answer that is $1+x$, in the given solution it is $1$, so it can't tbe the case that $f(x) = \sum_{n=0}^{\infty}(n+1)(x^n + x^{n+1})$, right? Wrong, because it can very well be that the "sums" (i.e. the series) are equal even though the first term isn't. The same way $1+2+3=2+1+3$ even though $1\neq 2$. $\endgroup$
    – Git Gud
    Feb 14, 2016 at 11:06
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    $\begingroup$ for $x=-1$, what you are doing is addition of two oscillating series which turns out to be zero. It is exact same thing like addition of $-1+1-1+1-1+1-1+...$ and $1-1+1-1+1-1+1-...$ becomes zero. $\endgroup$ Feb 14, 2016 at 11:07

2 Answers 2

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You did not end up with power series yet: \begin{align*} \sum_{n=0}^{\infty} (n+1)x^n+\sum_{n=0}^{\infty} (n+1) x^{n+1} &=\sum_{n=0}^{\infty} (n+1)x^n+\sum_{n=1}^{\infty} n x^{n}\\ &=\sum_{n=1}^{\infty} (2n+1) x^{n} + 1\\ &=\sum_{n=0}^{\infty} (2n+1) x^{n}\\ \end{align*}

From here you can derive correct radius of convergence.

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Hint: In order to find the power series expansion around $x=0$ you could also use the binomial series representation

\begin{align*} (1+x)^\alpha=\sum_{n=0}^{\infty}\binom{\alpha}{n}x^n\qquad\qquad \alpha\in\mathbb{C}, |x|<1 \end{align*}

We obtain \begin{align*} \frac{1+x}{(1-x)^2}&=(1+x)\sum_{n=0}^{\infty}\binom{-2}{n}(-x)^n\tag{1}\\ &=(1+x)\sum_{n=0}^{\infty}\binom{n+1}{n}x^n\\ &=(1+x)\sum_{n=0}^{\infty}(n+1)x^n\\ &=\sum_{n=0}^{\infty}(n+1)x^n+\sum_{n=0}^{\infty}(n+1)x^{n+1}\tag{2}\\ &=\sum_{n=0}^{\infty}(n+1)x^n+\sum_{n=1}^{\infty}nx^{n}\\ &=\sum_{n=0}^{\infty}(2n+1)x^n \end{align*}

Comment:

  • In (1) we use the identity $\binom{-n}{k}=\binom{n+k-1}{k}(-1)^k$

  • In (2) we shift the index of the right sum by one

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