Sum - geometric series

Question

I have series a + 2(1-a)a + 3 (1-a) (1-a)a + 4 (1-a)(1-a)a + … Does this series have a common ratio? How can I calculate sum and average?

Answer 1

Assuming you mean $4(1-a)(1-a)(1-a)a$ for your last-listed term, here's a way we can go:

First, observe that each term can be written as $$-a\cdot\frac{d}{da}\left[(1-a)^n\right]$$ for some positive integer $n,$ so that our series can be rewritten in the form $$-a\sum_{n=1}^\infty\frac{d}{da}\left[(1-a)^n\right],$$ and in fact, as $$-a\sum_{n=0}^\infty\frac{d}{da}\left[(1-a)^n\right].$$ Do you see why? Now, by power series derivative properties, we may rewrite that as $$-a\frac{d}{da}\left[\sum_{n=0}^\infty(1-a)^n\right].$$ Using the closed form of geometric series sums, we have for $|1-a|<1$ that $$\sum_{n=0}^\infty(1-a)^n=\frac{1}{1-(1-a)}=\frac{1}{a},$$ and so our series becomes $$-a\frac{d}{da}\left[\frac1a\right]=-a\cdot-\frac1{a^2}=\frac1a.$$

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As for the common ratio question: No. You can verify this by seeing that the second term is $2(1-a)$ times the first, while the third term is $\frac32(1-a)$ times the second. This can only be equal if $a=1,$ in which case our series becomes $1+0+0+0+0....$

As for average: I'm not at all sure what you mean.

Answer 2

We have:

$$a+ \left\{ \begin{aligned} (1-a) + (1-a)^2 + (1-a)^3 + \cdots \\ (1-a) + (1-a)^2 + (1-a)^3 + \cdots \\ (1-a)^2 + (1-a)^3 + \cdots \\ (1-a)^3 + \cdots \\ \end{aligned} \right. $$

This is: $$\left(a + \frac{1-a}{1-(1-a)} \right) + \frac{1-a}{1-(1-a)}+\frac{(1-a)^2}{1-(1-a)} + \frac{(1-a)^3}{1-(1-a)} + \cdots $$

which is another geometric series:

$$\left(a + \frac{1-a}{1-(1-a)} \right) + \frac{1}{a} \cdot\frac{1-a}{1-(1-a)}$$ $$=a + \frac{1-a}{a} + \frac{1}{a} \cdot \frac{1-a}{a}$$ $$= \frac{(a^2)+(a-a^2)+(1-a^2)}{a^2}$$ $$= \frac{-a^2+a+1}{a^2}$$ $$=-1 + \frac{1}{a} + \frac{1}{a^2}$$