I know that $\mathbb{E}(sup_n|X_n|)=\infty$ if $X_n=\frac{2^n}n\cdot\mathbf 1_{(1/2^{n+1},1/2^n)}$. However I am not sure how this can be evaluated explicitly. The probability space is $Ω=[0,1]$ endowed with its Borel sigma-algrebra and Lebesgue measure.
I get that
$\begin{align} \displaystyle \mathbb{E}(sup_n|X_n|) &=\int sup_n |X_n| d\mathbb{P} \\ &=\int sup_n \Big( \frac{2^n}{n}\cdot\mathbf 1_{(1/2^{n+1},1/2^n)}\Big) d\mu \\ \end{align}$
but not sure how this would evaluated.
Observe that $$ \sum_{k=0}^\infty \mathbf{1}_{(1/2^{k+1},1/2^k)}=\mathbf{1}_{[0,1]},\quad\lambda\, \text{- a.s.} $$ since they agree except in countably many points. Since $X_n$ is non-negative, $|X_n|=X_n$, and $$ \int_{[0,1]}\sup_n |X_n|\,\mathrm d\lambda=\int_{[0,1]} \sup_n X_n\mathbf{1}_{[0,1]}\,\mathrm d\lambda=\sum_{k=0}^\infty\int_{[0,1]}\sup_n X_n\mathbf{1}_{(1/2^{k+1},1/2^k)}\,\mathrm d\lambda. $$ But $\sup\limits_n X_n=\frac{2^k}{k}$ on the set $(1/2^{k+1},1/2^k)$ and hence $$ \begin{align} \int_{[0,1]}\sup_n |X_n|\,\mathrm d\lambda&=\sum_{k=0}^\infty \int_{[0,1]} \frac{2^k}{k}\mathbf{1}_{(1/2^{k+1},1/2^k)}\,\mathrm d\lambda\\ &=\sum_{k=0}^\infty \frac{2^k}{k}\left(\frac{1}{2^k}-\frac{1}{2^{k+1}}\right)=\infty. \end{align} $$