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I'm having a lot of trouble understanding a concept and was hoping someone could explain it to me. My professor did an example in class today, I will write out exactly what he did. We were asked to find the power series expansion of the following function about 0.

$$f(x)= \frac{x^2+x}{(1-x)^3} $$

So we started with the geometric series.

$$(1-x)^{-1} = \sum_{n=0}^\infty x^n $$

Differentiating once:

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

At this point, I understand that the first term in the above series is 0, so we shift the series to start with 1 and remove the first 0 term to get:

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

So at this point, he differentiates again:

$$2(1-x)^{-3} = \sum_{n=1}^\infty n(n-1)x^{n-2} $$

Dividing by two to get:

$$(1-x)^{-3} = \sum_{n=1}^\infty \frac {n(n-1)x^{n-2}}{2} $$

To make it of the form of the function at the start:

$$(x^2+x)(1-x)^{-3} = (x^3+x)\sum_{n=1}^\infty \frac {n(n-1)x^{n-2}}{2} $$

So he multiplied out the terms on the right to get:

$$(x^2+x)(1-x)^{-3} = \sum_{n=1}^\infty \frac {n(n-1)x^{n}}{2} + \sum_{n=1}^\infty \frac {n(n-1)x^{n-1}}{2} $$

So he then said, to get both of the series to have the term x to the power n, he shifts the very last series to make it start at 0, which I get:

$$(x^2+x)(1-x)^{-3} = \sum_{n=1}^\infty \frac {n(n-1)x^{n}}{2} + \sum_{n=0}^\infty \frac {n(n-1)x^{n}}{2} $$

Now here is where my question comes in,my professor said to get both series to start at the same number so you can add them, you can simply just shift the one that starts with 1 to start with 0, and it doesn't affect the terms inside the series:

$$(x^2+x)(1-x)^{-3} = \sum_{n=0}^\infty \frac {n(n-1)x^{n}}{2} + \sum_{n=0}^\infty \frac {n(n-1)x^{n}}{2} $$

I really don't get this. In all the other cases, when we shifted the series, it was either because we removed the first term which was 0, or we shifted it by adding or removing an n term. How can we just "shift it without affecting what is inside the series" If someone could explain this I would really be grateful! Thank you!

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    The difference between $\sum_{n=1}^\infty a_n$ and $\sum_{n = 0}^\infty a_n$ is $a_0$. And in your case $a_0$ is just $0$. – 2017-02-03
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    You are effectively saying that a quantity is unchanged by adding 0 to it; indeed the first term of the series indexed at $1$ is $\frac{n(n-1)x^n}{2}$ and at that quantity is 0 at both n=0 and n=1. – 2017-02-03
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    @rubik Hmm but I was told that if the first term in the series is 0 you should shift the series so the first term is not 0? That is why I shifted the series initially after differentiating once? Is this not a definite rule that I have to follow? Is making the series start with 0 not wrong? – 2017-02-03
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    @JohnMartin I am just unsure of why I had to shift the initial series after differentiating once if having the series start with 0 is not an issue? – 2017-02-03

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The first & seventh equations leave me confused ... $x^3$ or $x^2$ ? Was your Professor aiming for the nice formula \begin{eqnarray*} \frac{(x^2+x)}{(1-x)^{3}} = \sum_{n=0}^\infty n^2 x^{n}. \end{eqnarray*} When he shifted the formula, did you copy it down correctly ? \begin{eqnarray*} \sum_{n=0}^\infty \frac{n(n-1)}{2} x^{n-1} = \sum_{n=0}^\infty \frac{n(n+1)}{2} x^{n}. \end{eqnarray*} The start value of the sums does need remarking upon because some terms that are zero are being included in order to make the algebra workout later.

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    Hi! Sorry I changed my errors, it is supposed to be x to the power of 3. Yes the formula you wrote is the formula he got in the end. Also in your equation that you wrote is the start value supposed to change from 1 to 0? Or is it supposed to be 0 to 0? – 2017-02-03