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Suppose $$F(x)=\sum_{n \geq 0}{f_nx^n}=\frac{x}{1-3x}$$ We are asked to find an explicit formula for $f_n$. My working is , since$$\frac{1}{1-3x}=\sum_{n \geq 0}{3^nx^n} \\ \frac{x}{1-3x}=\sum_{n \geq 0}{3^nx^{n+1}}= \sum_{n \geq 0}{3^{n-1}x^n}$$ So we know that $f_n=3^{n-1}$ . I would like to ask when we do 'shifting' , wouldn't we also need to shift the index ? Like in this case , why we do not shift the index from $n \geq 0$ to $n \geq 1$ ?

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up vote 2 down vote accepted

You need to shift the index to $n\ge 1$. (If you don't do it, comparing the coefficient of $x^0$ - i.e. plugging $x=0$ - gives contradiction). The reasoning is as follows:

In $\sum_{n \geq 0}{3^nx^{n+1}}$ you did the substitution $m=n+1$ (equivalent to $n=m-1$), and so it becomes $\sum_{m-1 \geq 0}{3^{m-1}x^{m}}=\sum_{m \geq 1}{3^{m-1}x^{m}}$.

This is not very different from doing a variable substitution in an integral and then changing the limits of the integrals.

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Yes, you do need to shift the index - what you've written is incorrect. You should have $$\frac{x}{1-3x}=x\sum_{n\geq 0}3^n x^n=\sum_{n\geq 0} 3^nx^{n+1}=0x^0+3^0x^1+3^1x^2+\cdots=\sum_{n\geq 0} f_nx^n$$ where $$f_n=\begin{cases} 0 & \text{ if }n=0,\\ 3^{n-1} & \text{ if }n\geq 1.\end{cases}$$

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By shifting I think you mean $$\sum_{n \geq 0}{3^nx^{n+1}}= \sum_{n \geq 0}{3^{n-1}x^n}$$ Is this valid? The left hand side is $$ x^{1}+3x^2+9x^3+....$$ while the right hand side is $$ \frac13+x^{1}+3x^2+9x^3+....$$ As you can see they are not equal. What is valid is the following $$\sum_{n \geq 0}{3^nx^{n+1}}= \sum_{n \geq 1}{3^{n-1}x^n}=\sum_{n \geq 0}{3^{n-1}x^n}-\frac13$$ It can be easily seen that $f_0=0$ and $f_n=3^{n-1}$, $n\ge 1$

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$$ \sum_{n\geq 0}3^nx^{n+1}=\sum_{n\geq 1}3^{n-1}x^{n}=\sum_{n\geq 0}f_n x^n, $$ with $f_0=0$ and $f_n=3^{n-1}$ for $n\geq 1$.

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