It is well know that the convergence in distributions does not necessarily imply convergence in expectation, but implies convergence in expectation of bounded continuous functions.


Let $\{X\_n\}$ be a sequence of random variables that converge in distribution to $X$. I would like to ask two examples as follows.


1. An example such that $\mathbb{E}[X\_n]$ does not convergence to $\mathbb{E}[X]$.
2. An example such that $\mathbb{E}[X\_n^k]$ does not convergence to $\mathbb{E}[X^k]$ for all $k = 1, 2,\ldots$. That is, the sequence does not converge in all the moments, not just a or a few fixed moments. Suppose that all their moments exist.



Examples of convergence in distribution not implying convergence in moments
Let $P(X\_{n} = n) = \frac{1}{n}$ and $P(X\_{n} = 0) = 1 - \frac{1}{n}$. Then $X\_{n}$ converges in distribution to $X=0$, but $\mathbb{E}(X\_{n}^{k}) = \frac{1}{n} n^{k} + 0 = n^{k-1} \not\xrightarrow{n\to\infty} 0$ for each $k \in \mathbb{N}$.


**UPDATE:**
If you prefer continuous random variables on $\mathbb{R}$, you can "smoothen out" the previous example:


Let $X \sim N(0,1)$ and $X\_{n}$ have the probability density
$$
\rho\_{X\_{n}}(x)
=
\frac{1}{n} \cdot \frac{1}{\sqrt{2\pi}}\, \exp(-(x-n)^2/2) +
\left(1-\frac{1}{n}\right) \cdot \frac{1}{\sqrt{2\pi}}\, \exp(-x^2/2).
$$
Then $X\_{n} \stackrel{\mathrm{d}}{\to} X$.


Now denote the moments of $X$ by $A\_{k} := \mathbb{E}(X^{k})$ and, for $m \geq 0$, let $A\_{k,m} := \mathbb{E}((X+m)^{k}) \geq A\_{k} + m^k$.
It follows for each $k\in\mathbb{N}$:
\begin{align\*}
\mathbb{E}(X\_{n}^{k})
&=
\int\_{\mathbb{R}} x^{k} \rho\_{X\_{n}}(x)\, \mathrm dx
\\
&=
\frac{1}{n} A\_{k,n} + \left(1-\frac{1}{n}\right) A\_{k}
\\
&\geq
\frac{1}{n} (A\_{k}+n^{k}) + \left(1-\frac{1}{n}\right) A\_{k}
\\
&=
n^{k-1} + A\_{k}
\\
&\not\xrightarrow{n\to\infty}
A\_{k}.
\end{align\*}


