$(X\_k)\_k$ is a sequence of independent r.v uniformly bounded by $c.$ If $\sum\_{k}X\_k$ converges a.s then $\sum\_{k}E[X\_k]$ converges.


The above is proved using the following inequality ($X\_k$ should be centered):
$$P\left(\max\_{p \leq k}|Y\_p|>\epsilon\right) \geq 1-\frac{(\epsilon+c)^2}{E[Y^2\_k]} \ \ \ \ \ \ \ \ \ \ \ \ \ (1) $$ where $Y\_k=\sum\_{p=1}^kX\_k.$


If $X\_k$ are not centered then they consider independent copy of $X\_k.$


***The objective is to find a way to solve the problem without the use of symmetric random variable***, after searching, the following inequality (it seems it's a generalization of $(1)$) seems to solve the problem: $$P\left(\max\_{1 \leq p \leq k}|Y\_p| >\epsilon\right) \geq 1-4\frac{(\epsilon+\max\_{1 \leq p \leq k } |X\_p-E[X\_p]|)^2}{\operatorname{Var}(Y\_k)} \ \ \ \ \ \ \ \ (2)$$


Once $(2)$ is proved we can see that $\max\_{1 \leq p \leq k}|X\_p-E[X\_p]| \leq 2c$


How to prove $(2)$ ? (Of course the natural way to prove it is to consider $E\_p:=\{|Y\_1| \leq \epsilon,...,|Y\_{p-1}| \leq \epsilon,|Y\_p|>\epsilon\}$).


Any other inequalities to solve the problem are welcomed!



$P(\max_{1 \leq p \leq k}|Y_p| >\epsilon) \geq 1-4\frac{(\epsilon+\max_{1 \leq p \leq k } |X_p-E[X_p]|)^2}{\operatorname{Var}(Y_k)}$
The statement



> 
> If $(X\_k)$ is a sequence of independent r.v.'s uniformly bounded in absolute value by some real $c$ such that $\sum\_{k}X\_k$ converges a.s., then $\sum\_{k}EX\_k$ converges
> 
> 
> 


follows almost immediately from the Hoffmann–Jørgensen inequality
$$EM^2\le2.4^2c^2+32t\_0^2$$
(see e.g. [Proposition 6.8](https://link.springer.com/book/10.1007/978-3-642-20212-4)),
where $M:=\sup\_k|S\_k|$, $S\_k:=\sum\_{j=1}^k X\_j$, and $t\_0>0$ is such that $P(M>t\_0)<1/32$. Indeed, since $S\_\infty$ is finite a.s., we have $M<\infty$ a.s. and hence we can assume that $t\_0<\infty$, so that $Var\, S\_n\le ES\_n^2\le EM^2\le2.4^2c^2+32t\_0^2<\infty$ for all natural $n$.


Now Kolmogorov's [two-series theorem](https://en.wikipedia.org/wiki/Kolmogorov%27s_two-series_theorem) implies that $\sum\_k(X\_k-EX\_k)$ converges a.s. Since $\sum\_k X\_k$ converges a.s., we conclude that $\sum\_k EX\_k$ converges, as desired.


**Extended comment:** As the OP noted in a comment, inequality (2) does appear, as Satz III, in [Kolmogorov's paper](https://link.springer.com/content/pdf/10.1007/BF01459098.pdf) (of 1928, in German). However, apparently there is a mistake in the proof of this inequality in that paper. Namely, with standard modern notations: $E$ in place of $\mathfrak D$ (the expectation symbol in that paper), $A\_k$ in place of $\mathfrak f\_k$ (a certain event), and $EX1\_{A\_k}$ in place of $\mathfrak D\_{\mathfrak f\_k}(X)$, the equality in line 10 on p. 312 of Kolmogorov's paper becomes
$$|t\_k-Et\_k1\_{A\_k}|=|s\_k-Es\_k1\_{A\_k}|,$$
which was apparently thought of as the consequence of the equality
$$t\_k-Et\_k1\_{A\_k}=s\_k-Es\_k1\_{A\_k};\tag{$\*$}$$
here, according to p. 310 of Kolmogorov's paper, $t\_k=s\_k-Es\_k$. However, actually
$$t\_k-Et\_k1\_{A\_k}=s\_k-Es\_k1\_{A\_k}-Es\_k(1-P(A\_k)).$$
So, ($\*$) holds only if $Es\_k=0$ or $P(A\_k)=1$, which latter is not true in general. Thus, the proof seems to only work if $Es\_k=0$ for all $k$. This is the first time in my life where I seem to see a mistake made by Kolmogorov!


It appears that this mistake was (tacitly?) corrected in later publications by symmetrization (which obviously provides for the zero-mean condition). If that correction was made indeed tacitly, it appears to be a disservice to people interested in this matter, as the question on this webpage shows.


Anyhow, the Hoffmann–Jørgensen inequality seems to be a more effective and more transparent way to achieve the desired result, as shown above in this answer.


