TITLE: Simple proof of a theorem on convergence of series
QUESTION [5 upvotes]: Let $p_j\ge0,\ j=1,2,3,\dots,$ and suppose $\sum_j p_j=1.$  Is there a simple proof that $$\sum_{j=1}^\infty{jp_j}\tag{1}$$ converges?  My question arises from the answer to this question.  Consider a Markov chain with state space $\{1,2,3,\dots\}.$  If the chain is is state $1,$ it transitions to state $j$ with probability $p_j.$  If it is in state $j>1$ then it always transitions to state $j-1$.  The chain is irreducible and aperiodic, so it has a unique stationary distribution.  The sum $(1)$ arises in computing the stationary probabilities, so it must converge.
I've been trying unsuccessfully to find a more direct proof.  There isn't any way to apply standard tests (root test, ratio test, Gauss's test) and I haven't any other ideas.  (It's equivalent to the statement that if N is a random variable that takes positive integer values, then $E(N)$ exists, but I don't see how that helps.  In fact, my intuition would be that this statement is false.)
EDIT 
It has been amply shown that the statement is false.  I would like to know the error in the linked question.

REPLY [4 votes]: Another counterexample can be derived from the St. Petersburg paradox. Suppose that $p_j=\frac1j$ if $j$ is a power of $2$, and $0$ otherwise. Then $\sum p_j = \sum 2^{-k}=1$, but $jp_j=1$ whenever $j$ is a power of $2$, and thus $\sum jp_j$ diverges.