formula for infinite sum of a geometric series with increasing term
I'm looking for the Expectation of the discrete random variable X, E[X], with pmf: $$p(x)=(\frac 16)^{x+1}, x=0,1,2,3...$$
so what I tried is as follows... $$E[X]= \sum_{0}^\infty xp(x) =$$
so then $$=\sum_0^\infty x(\frac 16)(\frac 16)^x = \frac 16\sum_{0}^\infty x(\frac 16)^x $$
which is $$ \frac 16 [0(\frac 16)^0+1(\frac 16)^1+2(\frac16)^2+3(\frac 16)^3+...] $$
so if we were to use the rule that: $$ \sum_0^\infty ar^x = \frac {a}{(1-r)} $$ when |r|<1.
Then it seems like the difference between that formula and my problem is the increasing coefficient on the (1/6)^x...
My math book (which doesn't really say anything more about it)... states that "there is a general increasing geometric series relation which is $$1 + 2r + 3r^2 + 4r^3+...= \frac {1}{(1-r)^2} $$
is that what I need to know? and if so, can someone please show me why? Thanks!
$\endgroup$ 22 Answers
$\begingroup$This "general relation" follows from calculus where you learn about Maclaurin series
In particular, it talks about how since
$$\sum_{n=0}^\infty x^n = {1\over 1-x},\quad |x|<1$$
We can take derivatives of both sides and get
$$\sum_{n=0}^\infty{d\over dx}(x^n)={d\over dx}\left(\sum_{n=0}^\infty x^n\right)={d\over dx}\left({1\over 1-x}\right)$$
therefore
$$\sum_{n=0}^\infty nx^{n-1}={1\over (1-x)^2}$$
In your case you use $x$ instead of $n$ and ${1\over 6}$ instead of $x$, but it amounts to the same thing, just using different letters.
So you are trying to solve
$$\sum_{n=0}^\infty n\left({1\over 6}\right)^{n+1}={1\over 36}\sum_{n=0}^\infty n\left({1\over 6}\right)^{n-1}={1\over 36}\cdot{1\over \left(1-{1\over 6}\right)^2}={1\over (6-1)^2}={1\over 25}$$
Addendum
I almost forgot, you should be careful to renormalize your mass function. Note that
$$\sum_{n=0}^\infty\left({1\over 6}\right)^{n+1}={1\over 6}\cdot {1\over 1-{1\over 6}}={1\over 5}\ne 1.$$
This is a problem, because all the probabilities have to add up to $1$. So either there's a missing piece which has the rest of the probability, or you really mean that the pmf is
$$P(X=x)=5\cdot \left({1\over 6}\right)^{x+1}$$
in which case the final answer changes from ${1\over 25}$ to ${1\over 5}$. Either way, something has to change, as without a true probability measure, the answer doesn't make sense in context.
$\endgroup$ 6 $\begingroup$Try multiplying out $(1+r+r^2+\cdots)^2$. The coefficient of $r^k$ will be the number of ways of adding two nonnegative integers to get $k$, i.e. $k+1$.
$\endgroup$ 1
