How to find a sequence of random variables of infinite expectation that converges to zero? I want to find an example such that $X_n\rightarrow0$ as $n\rightarrow \infty$ while $E(X_n)=\infty$.
I am thinking of making use the divergence of $\sum\frac{1}{n}$, but fail to find suitable $X_n$ and $\Pr(X_n=k)$.
Can you show me some examples or give me some hints?
 A: To make a series of random variables $X_n$ approach $0$ in probability, we want the chance that $X_n$ is not $0$ to become vanishingly small.  At the same time, to ensure the expectations of each of the $X_n$ diverge, we need to leave some probability out in the tails (that is, where the values of the $X_n$ are large enough to overwhelm the small probabilities that $X_n$ is nonzero).
To do this, let's consider discrete distributions supported on the natural numbers.  Their probability mass functions $f_n(k)$, $k=0, 1, 2, \ldots,$ need to have these properties:


*

*The limiting value of $f_n(0)$ must be $1$ as $n\to\infty$.

*Every expectation, equal to $\sum_{k=0} k f_n(k)$, must diverge to $\infty$.
The question suggests we make use of the fact that the Harmonic series $\sum_{k=1} 1/k$ diverges.  Comparing that to (2) would have us try
$$k f_n(k) = 1/k,$$
at least for sufficiently large $k$ ($k=0$ would be a problem, of course).  So, solving these equations for $k$, let's see what happens when $f(k)$ is made proportional to $1/k^2$, $k \gt 0$.  For this to define a probability distribution, we would need the masses to be nonnegative and sum to unity.  But $\sum_{k=1} 1/k^2 = \pi^2/6$.  That's exactly what's needed to normalize $f$.  Now as $n$ increases, take a little mass out of this positive part of $f$ and move it to $0$.  An easy way to accomplish this would be to cut down the positive mass by $1/n$. Whence, defining
$$f_n(k) = \frac{1}{n}\frac{6}{\pi^2}\frac{1}{k^2} = \frac{6}{n\pi^2 k^2}, \ k \gt 0;\quad f_n(0) = 1 - \frac{1}{n}$$
will do it.
