Sequence from generating function $\frac{1}{(1 - \frac{x}{3})^2}$

Question

I know that for

$$ \frac{1}{1 - \frac{x}{3}} $$

sequence would be $a_n = \frac{1}{3^n}$ for $n \geq 0$ ($\sum_{n \geq 0} \frac{1}{3^n} x^n$ ).

How should I approach with that power of two?

Answer 1

$$\frac{1}{(1-\frac{x}{3})^2}=\frac{1}{1-\frac{x}{3}}\cdot\frac{1}{1-\frac{x}{3}}$$ $$=\sum\limits_{n=0}^\infty(\frac{x}{3})^n\cdot\sum\limits_{n=0}^\infty(\frac{x}{3})^n $$ $$=(1+\frac{x}{3}+(\frac{x}{3})^2+\ldots)(1+\frac{x}{3}+(\frac{x}{3})^2+\ldots).$$ Now we want a single sum in the form of $\sum a_nx^n$, and so we're going to find all of the terms resulting from the multiplication above that correspond to $x^0, x^1, x^2,...$ and match those coefficients to $a_0, a_1, a_2,...$. For example, the only way to get a constant term is from multiplying the first terms of both sequences (1) together, so $a_0=1$. Then, there are two ways to get $x^1$, either from multiplying the 1 in the first sequence to $\frac{x}{3}$ in the second or vice versa, so we get $a_1=2\cdot\frac{1}{3}$. In general, when we add 1 to our index $n$, there is one more way to get an $x^n$ term. And so $a_n=(n+1)\cdot(\frac{1}{3})^n$, for $n\ge0$. Thus, your series would be: $$\sum\limits_{n=0}^\infty (n+1)\frac{x^n}{3^n}.$$

Answer 2

In general, by computing the Taylor series expansion.

In this case it happens to work to use $\frac{1}{(1-z)^2}=\frac{\partial}{\partial z}\frac{1}{1-z}$.

But a more widely applicable bit of knowledge is that if $f(z)=\sum_{n=0}^\infty a_nz^n$, then

$$\frac{1}{1-z}\,f(z)=\sum_{n=0}^\infty\left(\sum_{k=0}^n a_k\right)z^k.$$

And $\frac{1}{(1-z)^2}=\frac{1}{1-z}\frac{1}{1-z}$.

Lastly, set $z=\frac{x}{3}$ for a cleaner computation.

Answer 3

Same, just use the binomial coefficient $\binom{-2}{k}$. If you feel scared, just set $\alpha=-2$ and do the Binomial expansion. You should get something like $(-1)^k k(k+1)$ I think.

Answer 4

For yet another answer,

$$ \frac1{1-x} = \sum_{n=0}^\infty x^n,$$

and

$$ \frac1{1-\frac13 x} = \sum_{n=0}^\infty \left(\frac13\right)^n x^n.$$

Differentiating we have

$$ \frac{\frac13}{\left(1-\frac13 x\right)^2} = \sum_{n=0}^\infty(n+1)\left(\frac13\right)^{n+1}x^n. $$

Hence $$\frac1{\left(1-\frac13 x\right)^2} = \sum_{n=0}^\infty(n+1)\left(\frac13\right)^nx^n. $$