Expected Value of variate based on RNG A random number generator produces independent random variates $x_0, x_1, x_2, ...$ drawn uniformly from $[0,1]$, stopping at the first $x_T$ which is strictly less than $x_0$. Prove that the Expected Value of T is infinite.
So I've got that $E[T] = P[T=1] + 2P[T=2] + 3P[T=3] +  ...$
I'm not sure how to show this is infinite because don't we have that $P[T=k] \to 0$ as $k$ goes to infinity?
Would appreciate any tips, thanks in advance.
 A: Let's begin with calculating the following conditional probability:
$$P(T=n\mid X_0=x)=x(1-x)^{n-1}=\sum_{k=0}^{n-1}(-1)^{n-1-k}x^{n-k}.$$
Therefore 
$$P(T=n)=\sum_{k=0}^{n-1}(-1)^{n-1-k}\int_0^1x^{n-k} dx=\sum_{k=0}^{n-1}(-1)^{n-1-k}\frac{1}{n-k+1}$$
and
$$E[T]=\sum_{n=1}^{\infty}n\sum_{k=0}^{n-1}(-1)^{n-1-k}\frac{1}{n-k+1}\geq\sum_{n=1}^{\infty}\frac n{n+1}\sum_{k=0}^{n-1}(-1)^{n-1-k}=$$
$$=\sum_{\text{for odd n's}}\frac n{n+1}=\infty.$$
(Note that $\sum_{k=0}^{n-1}(-1)^{n-1-k}$ is $0$ for even $n$'s and is $1$ for odd $n$'s.)
A: I'd like to add another solution (a bit more general, doesn't rely on a uniform distribution).
The probability $P\{T>k\}=P\{x_1\ge x_0,\dots,x_k\ge x_0\}$. Denote $A_i=\{x_0\ge x_i,\dots, x_k\ge x_i\}$, $i=0$, ..., $k$. Since the $x_0$, ..., $x_k$ are independent and identically distributed, $P(A_i)$ are the same for all $i$. Also, $A_0\cup\dots\cup A_k=\Omega$, so $P(A_0\cup\dots\cup A_k)=1$, and $P(A_i)\ge1/(k+1)$ for all $i$ (we are interested only in $P(A_0)$). To summarize: $P\{T\ge k+1\}=P\{T>k\}=P(A_0)\ge1/(k+1)$.
The expectation $$E(T)=\sum_{k=1}^\infty kP\{T=k\}=\sum_{k=1}^\infty P\{T\ge k\}\ge\sum_{k=1}^\infty\frac{1}{k}=+\infty.$$
