A.S. Convergence of Sum of Independent Random Variables I am trying to prove the following: 
Let $\xi_{1},\xi_{2},...$ be independent random variables with non-negative values. Prove that the series$\sum_{i=1}^{\infty} \xi_{i}$ converges almost surely if and only if $\sum_{i=1}^{\infty} E( \frac{\xi_{i}}{1+\xi_{i}})<\infty$.
Here's my line of thought so far: Since the r.v.'s are non-negative by the monotone convergence theorem we have $\sum_{i=1}^{\infty} E(\frac{\xi_{i}}{1+\xi_{i}}) = E\sum_{i=1}^{\infty} \frac{\xi_{i}}{1+\xi_{i}}$.  If $\sum_{i=1}^{\infty} \xi_{i}<\infty$, then clearly so is $\sum_{i=1}^{\infty} \frac{\xi_{i}}{1+\xi_{i}}$ and hence $E\sum_{i=1}^{\infty} \frac{\xi_{i}}{1+\xi_{i}} < \infty$.  
Now to show the reverse we assume $\sum_{i=1}^{\infty} E(\frac{\xi_{i}}{1+\xi_{i}}) < \infty$ and want to show that $\sum_{i=1}^{\infty} \xi_{i}<\infty$. By the above reasoning $E\sum_{i=1}^{\infty} \frac{\xi_{i}}{1+\xi_{i}} < \infty$ and any r.v. with finite expectation is also finite so $\sum_{i=1}^{\infty} \frac{\xi_{i}}{1+\xi_{i}} < \infty$.  This is where I am stuck.  I am assuming I need to apply the fact that the r.v.'s are independent at some point, but I'm not sure where...
 A: 
Lemma 1 (Kolmogorov’s three-series theorem)
  Let $\left\{ X_n \, | \, n\geqslant 1 \right\}$  be a sequence of independent random variables. Let $A>0$  be a real number and let $X_n^A = X_n 1_{\left\{ \left | X_n \right |\leqslant A\right\}}$, where $1_{\left\{\dots\right\}}$ denotes the indicator function of a set.
  Let us define three conditions:
  \begin{align*}
& (K_1)\, \sum_{n=1}^{\infty}P\left ( \left | X_n \right |> A \right )<+\infty \, ; \\
& (K_2)\, \sum_{n=1}^{\infty}E\left ( X_n^A \right ) \text{ converges}\, ;  \\
& (K_3)\, \sum_{n=1}^{\infty}Var\left ( X_n^A \right ) <+\infty \, .   
\end{align*}
  Then the following three propositions are equivalent:
  \begin{align*}
& (P_1)\, \sum X_n \text{ converges almost surely}\, ; \\
& (P_2)\, \forall A>0\, ,  \text{ $(K_1)$ $(K_2)$ and $(K_3)$ hold true }  \, ;     \\
& (P_3)\, \exists A>0\, ,  \text{ $(K_1)$ $(K_2)$ and $(K_3)$ hold true for that }  A \, .
\end{align*}         



Lemma 2
  Let $X$ be a random varible. Let $A>0$ be a real number and $Z= \left | X \right |1_{\left \{ \left | X \right |\leqslant A \right \}}$ . Then we have
  $$  \frac{1}{2A} E\left (Z\right )+\frac{1}{2}P\left ( \left | X \right | > A  \right ) \leqslant E\left ( \frac{\left | X \right |}{\left | X \right |+A} \right ) \leqslant  \frac{1}{A} E\left ( Z \right )+P\left ( \left | X \right | > A  \right )  $$
  Proof:
  \begin{align*}
 &E\left ( \frac{\left | X \right |}{\left | X \right |+A} \right )=E\left ( \frac{\left | X \right |}{\left | X \right |+A} 1_{\left \{ \left | X \right |\leqslant A \right \}} \right )+E\left ( \frac{\left | X \right |}{\left | X \right |+A} 1_{\left \{ \left | X \right |> A \right \}} \right )  \\
 &\qquad\qquad\qquad \leqslant \frac{1}{A} E\left ( \left | X \right | 1_{\left \{ \left | X \right |\leqslant A \right \}} \right )+E\left ( 1_{\left \{ \left | X \right |> A \right \}} \right )  \\ 
 &\qquad\qquad\qquad =  \frac{1}{A} E\left (Z \right )+P\left ( \left | X \right | > A  \right ) 
\end{align*}
  \begin{align*}
 &E\left ( \frac{\left | X \right |}{\left | X \right |+A} \right )=E\left ( \frac{\left | X \right |}{\left | X \right |+A} 1_{\left \{ \left | X \right |\leqslant A \right \}} \right )+E\left ( \frac{\left | X \right |}{\left | X \right |+A} 1_{\left \{ \left | X \right |> A \right \}} \right )  \\
 &\qquad\qquad\qquad \geqslant  \frac{1}{2A} E\left ( \left | X \right | 1_{\left \{ \left | X \right |\leqslant A \right \}} \right )+\frac{1}{2} E\left ( 1_{\left \{ \left | X \right |> A \right \}} \right )  \\ 
 &\qquad\qquad\qquad =  \frac{1}{2A} E\left (Z \right )+\frac{1}{2}P\left ( \left | X \right | > A  \right ) 
\end{align*}


First of all, define $\xi_n^A=\xi_n 1_{\left\{  \xi_n \leqslant A\right\}}$ for $A>0$.  
(1)Necessity:
Now we assume that $\sum_{n=1}^{\infty}\xi_n$ converges almost surely. 
Let $A=1>0$, by lemma 1, we have
$$\sum_{n=1}^{\infty}P\left ( \xi_n> 1 \right )<+\infty \text{ and }\sum_{n=1}^{\infty}E\left ( \xi_n^1 \right )<+\infty .$$
For each $n$, by lemma 2, we have
$$ E\left ( \frac{\xi_n}{\xi_n+1} \right ) \leqslant  E\left ( \xi_n^1 \right )+P\left ( \xi_n > 1  \right ) .$$
Hence
$$\sum_{n=1}^{\infty}E\left ( \frac{\xi_n}{\xi_n+1} \right ) \leqslant  \sum_{n=1}^{\infty}E\left ( \xi_n^1 \right )+\sum_{n=1}^{\infty}P\left ( \xi_n > 1  \right )<+\infty .$$
(2)Sufficiency:
Now we assume that $\sum_{n=1}^{\infty}E\left ( \frac{\xi_n}{1+\xi_n} \right )<+\infty$. 
For each $n$, by lemma 2, we have
$$ E\left ( \frac{\xi_n}{\xi_n+1} \right ) \geqslant  \frac{1}{2}E\left ( \xi_n^1 \right )+\frac{1}{2}P\left ( \xi_n > 1  \right ) .$$
Hence we have
\begin{align*}
 &\sum_{n=1}^{\infty}P\left ( \xi_n > 1  \right )\leqslant 2\sum_{n=1}^{\infty}E\left ( \frac{\xi_n}{\xi_n+1} \right )<+\infty, \\
 &\sum_{n=1}^{\infty}E\left ( \xi_n^1 \right )\leqslant 2\sum_{n=1}^{\infty}E\left ( \frac{\xi_n}{\xi_n+1} \right )<+\infty. 
\end{align*}
Furthermore
\begin{align*}
 &\sum_{n=1}^{\infty}Var\left ( \xi_n^1 \right ) \leqslant \sum_{n=1}^{\infty}E\left [  \left( \xi_n^1 \right ) \cdot \left( \xi_n^1 \right ) \right ]   \\
 &\qquad\qquad\quad\,\, = \sum_{n=1}^{\infty}E \left [ \left(\xi_n\right) \cdot \left(\xi_n\right) \cdot 1_{\left\{  \xi_n \leqslant 1\right\}} \right ] \\
 &\qquad\qquad\quad\,\, \leqslant  \sum_{n=1}^{\infty}E \left [ \xi_n \cdot 1_{\left\{  \xi_n \leqslant 1\right\}} \right ] \\ 
 &\qquad\qquad\quad\,\, =\sum_{n=1}^{\infty}E\left ( \xi_n^1 \right )<+\infty. 
\end{align*}
Now (K1) (K2) and (K3) hold ture for $A=1$, so we can get the conclusion that $\sum_{n=1}^{\infty}\xi_n$ converges almost surely by lemma 1.
