Showing that the series of random variables $X_n / n$ converges almost surely under some conditions I'm working on a problem from Chow and Teicher's book on Probability Theory, page 123, #6(ii):
If $X_n, n\geq 1$ are i.i.d., $\mathcal{L_1}$ r.v.s, then $\sum (X_n / n)$ converges a.c. if $E|X_1|log^+ |X_1| <\infty$ and $EX_1 = 0$.
The most relevant theorem that I've been thinking of using is one that says if $X_n$ are i.i.d. and $\mathcal{L_p}$ for $0<p<2$, then $\sum \left(X_n / n^{1/p} - E\left(\dfrac{X_nI_{\{|X_n|\leq n^{1/p}\}}}{n^{1/p}}\right)\right)$ converges a.c.; since, in this case, $p =1$, it would suffice to show that the series $\sum E\left(\dfrac{X_nI_{\{|X_n|\leq n\}}}{n}\right)$ converges a.c. to complete the exercise. However, I'm having trouble incorporating the $E|X_1|log^+ |X_1| <\infty$ condition. I see how $EX_1 = 0$ implies that the summands of the series satisfy the following: $E\left(\dfrac{X_nI_{\{|X_n|\leq n\}}}{n}\right) = -E\left(\dfrac{X_nI_{\{|X_n|> n\}}}{n}\right)$, but unfortunately I've been thus far unable get anything resembling a logarithmic series by manipulating summands here.
Perhaps I'm missing something obvious? Any help is greatly appreciated. 
 A: Since the random variables are identically distributed, the problem will be solve if we prove that the series $\sum_{n\geqslant 1}\mathbb E[|X_0|\chi_{\{|X_0|\geqslant 2^n\}}]$ converges. 
This can be seen noticing that the assumption $\mathbb E[|X_0|\cdot|\log^+|X_0||$  finite implies the convergence of the series $\sum_{k\geqslant 1}k2^k\mu\{|X_0|\geqslant 2^k\}$ and $\mathbb E[|X_0|\chi_{\{|X_0|\geqslant 2^n\}}]$ can be expressed in terms of $\mu\{|X_0|\geqslant 2^k\}$.
Let us show the convergence of $\sum_{k\geqslant 1}k2^k\mu\{|X_0|\geqslant 2^k\}$. For a random variable $Y\geqslant 2$, the following equality holds 
$$\mathbb E[Y\log Y]=\int_2^{+\infty}(1+\log t)\mu\{Y\geqslant t\}\mathrm dt=\sum_{k=1}^{+\infty}\int_{2^k}^{2^{k+1}}(1+\log t)\mu\{Y\geqslant t\}\mathrm dt.$$
Then we use the bound $\int_{2^k}^{2^{k+1}}(1+\log t)\mu\{Y\geqslant t\}\mathrm dt\geqslant 2^k(k+1)\mu\{Y\geqslant 2^{k+1}\}$.
To conclude, notice that 
$$\mathbb E[Y\chi_{\{Y\geqslant R\}}]=\int_R^{+\infty}\mu\{Y\geqslant t\}\mathrm dt+R\mu\{Y\geqslant R\}.$$
A: Actually, there is a more elementary proof. We can indeed conclude once we show the convergence of $\sum_{n\geqslant 1}n^{-1}\mathbb E[|X_0|\chi_{\{|X_0|\geqslant n\}}]$. Define $A_k:=\{k\leqslant |X_0|\lt k+1\}$. Then, 
\begin{align}
\sum_{n\geqslant 1}n^{-1}\mathbb E[|X_0|\chi_{\{|X_0|\geqslant n\}}]&=
\sum_{n\geqslant 1}n^{-1}\sum_{k\geqslant n}\mathbb E[|X_0|\chi_{A_k}]\\
&=\sum_{k\geqslant 1}\sum_{n=1}^kn^{-1}\mathbb E[|X_0|\chi_{A_k}]\\
&\leqslant \sum_{k\geqslant 1}\frac 1{\log k}\mathbb E[|X_0|\cdot \log^+|X_0|\chi_{A_k}]\sum_{n=1}^kn^{-1}\\
&\leqslant \sum_{k\geqslant 1}\mathbb E[|X_0|\cdot \log^+|X_0|\chi_{A_k}],
\end{align}
and this series is convergent because $|X_0|\log^+|X_0|$ is integrable.
