Generating function of (-3)^n

Question

Given the sequence $a_n$, where $n$-th element is $a_n = (-3)^n$

I have the generating function $$A(t) = \sum_{n=0}^{\infty} (-3)^nt^n $$

The problem now is to simplify the obtained expression.

I did the following:

Let $B(t) = -3t$ be the generating function and $C(B(t)) = A(t) = \sum_{n=0}^{\infty} (-3)^nt^n $. Hence, the generating function $C(t) = \sum_{n=0}^{\infty} t^n = \frac{1}{1 - t}$

Therefore $$A(t) = C(B(t)) = C(-3t) = \frac{1}{1 + 3t}$$

Is this a possible way to solve this problem?

Your basic approach is correct, but $C(t)=\frac1{1-t}$, not $\frac{t}{1-t}$; $\frac{t}{1-t}=\sum_{i\ge 1}t^n$. Thus, the numerator of $A(t)$ should also be $1$. — Brian M. Scott, Oct 30 '15 at 20:16
@Brian M. ScottBy the way. I don't think why we consider that $|t| < 1$? — J.Exactor, Oct 30 '15 at 20:25
If you want the series to converge, you need to take $|-3t|=3|t|<1$, so you actually want $|t|<\frac13$. — Brian M. Scott, Oct 30 '15 at 20:26
@Brian M. Scott As a result, I have to replace $\frac{t}{1 - t}$ by $\frac{1}{1 - t}$ or to change the numerator? — J.Exactor, Oct 30 '15 at 20:29
Replace $\frac{t}{1-t}$ by $\frac1{1-t}$; you should then get the correct result for $A(t)$. — Brian M. Scott, Oct 30 '15 at 20:34

score 2 · Answer 1 · answered Oct 30 '15 at 20:43

2

You could simply notice that $\sum_{n=0}^\infty(-3)^nt^n$ is a convergent geometric series with ratio $-3t$ for $\|3t\|< 1$.

The closed form of geometric series with ratio $r$, where $\|r\| < 1$, is $$ \sum_{n=0}^\infty r^n = \frac{1}{1-r}. $$

answered Oct 30 '15 at 20:43

Myath

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Generating function of (-3)^n

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