How to find $\sum\limits_{j=1}^{\infty}j(1-p)^j$?

Question

How to prove $\sum\limits_{j=1}^{\infty}j(1-p)^j=\frac{1-p}{p^2}$?

Is there any general ways to think about this kind of problems?

Answer 1

If you know that $$\sum_{n=0}^{\infty} x^n=\frac {1}{1-x} \ ,|x| < 1$$ Then take the derivative and multiply by $x$ to get your sum.

Answer 2

$$S=\sum_{j=1}^{n} j(1-p)^j=1(1-p)+2(1-p)^2+3(1-p)^3+...+n(1-p)^n$$

Multiply by $1-p$

$$(1-p)S=1(1-p)^2+2(1-p)^3+3(1-p)^4+...+n(1-p)^{n+1}$$

$$S-(1-p)S=1(1-p)+1(1-p)^2+...+1(1-p)^n+n(1-p)^{n+1}$$

$$pS=(1-p)\frac{1-(1-p)^n}{1-(1-p)}+n(1-p)^{n+1}=$$

$$S=\left(\frac{1-p}{p}\right)\frac{1-(1-p)^n}{p}+\frac{n(1-p)^{n+1}}{p}$$

Now make $n \rightarrow \infty$. If you have $|1-p|<1$ then your sum will go to:

$$\left(\frac{1-p}{p^2}\right)$$

P.S: You can use the same idea for a sum like:

$$S=a_1g_1+a_2g_2+...+a_ng_n$$

where $a_n$ is an arithmetic sequence and $g_n$ is a geometric sequence.

Answer 3

You can also see this as a two geometric series when you write out the terms, so under assumption of convergence you can reorder and calculate (using geometric series formula where appropriate): \begin{align} \sum_{j=1}^{\infty}j(1-p)^j &= (1-p) + [(1-p)^2+(1-p)^2] + [(1-p)^3+(1-p)^3+(1-p)^3] + \dots \\ &= [(1-p) + (1-p)^2 + \dots] +[(1-p)^2 + (1-p)^3 + \dots] + \dots \\ &= \frac{1-p}{1-(1-p)} + \frac{(1-p)^2}{1-(1-p)} + \dots\\ &= \frac{1}{p} ((1-p) + (1-p)^2 + \dots)\\ &= \frac{1}{p} \left(\frac{1-p}{1-(1-p)}\right)\\ &= \frac{1-p}{p^2}\\ \end{align}