Rate of convergence when going from convergence in probability to convergence in the mean Consider positive and bounded random variables $(X_n)_{n\in\mathbb{N}}$. By Helly-Bray theorem (or dominated convergence theorem), we have 
$$
 ((X_n) \text{ converge in probability towards } 0) \implies ((X_n) \text{ converge in the mean towards } 0)
$$
Question: Assume that we know the rate of convergence in probability of $(X_n)$ towards 0. What is the rate of convergence in the mean of $(X_n)$ towards 0?
 A: A sequence $\left(X_n\right)_{n\geqslant 1}$ converges in probability to $0$ if and only if $$\delta_n:=\mathbb E\left[\frac{\left\lvert X_n\right\rvert    }  {1+\left\lvert X_n\right\rvert }   \right]\to 0.$$
Indeed, if $X_n\to 0$ in probability, then 
$$\mathbb E\left[\frac{\left\lvert X_n\right\rvert    }  {1+\left\lvert X_n\right\rvert }   \right]\leqslant\varepsilon +\mathbb E\left[\frac{\left\lvert X_n\right\rvert    }  {1+\left\lvert X_n\right\rvert }  \mathbf 1\left\{\left\lvert X_n\right\rvert\gt \varepsilon  \right\}         \right]  \leqslant\varepsilon +\mathbb P\left\{\left\lvert X_n\right\rvert\gt \varepsilon  \right\}.$$
The converse can be shown using monotonicity of $x\mapsto x/(1+x)$ for non-negative $x$.  
If there exists a constant $M$ such that $\left\lvert X_n\right\rvert\leqslant M$ almost surely, then 
$$\mathbb E\left\lvert X_n\right\rvert\leqslant \left(M+1\right)\delta_n.$$ 
Alternatively, denoting $\delta(n,\varepsilon) :=  \mathbb P\left\{\left\lvert X_n\right\rvert\gt \varepsilon  \right\}$, we have 
$$\mathbb E\left\lvert X_n\right\rvert\leqslant 
\varepsilon +\mathbb E\left[  \left\lvert X_n\right\rvert\mathbf 1\left\{\left\lvert X_n\right\rvert\gt \varepsilon  \right\}   \right]  \leqslant \varepsilon+M\delta(n,\varepsilon).         $$
