A different type of Law of Large Numbers Let $(X_i)_{i\in\mathbb{N}}$ be a sequence of i.i.d. real random variables (with finite variance if needed).
For $k>0$ fixed, I'm interested in estimating the quantity
$$ p_N(k):=\mathbb{P}(\exists\,i=1,\dots,N\text{ s.t. }|X_i-\mathbb{E}[X_i]|\geq k\,N) \;.$$
Using the Markov inequality, if $X_1$ has finite variance, it can be shown that
$\sup_{N\in\mathbb{N}} p_N(k)\to0$ as $k\to\infty$.
I would like to prove that, for $k$ large enough, $\sum_{N=1}^\infty p_N(k) <\infty$. Is it true?
 A: Note that $p_N(k)\leqslant N\,q_N(k)$, with $q_N(k)=P[|X-E[X]|\geqslant kN]$. Furthermore, the series $\sum\limits_NNq_N(k)$ converges if and only if $X^2$ is integrable.
On the other hand, $p_N(k)=1-(1-q_N(k))^N$ hence, if $Nq_N(k)\to0$, then $p_N(k)=\Theta(Nq_N(k))$.
Thus, for any positive $k$, the series $\sum\limits_Np_N(k)$ converges if and only if $X^2$ is integrable.
Edit: Quite generally, consider a nonnegative sequence $(a_n)$ and a nonnegative random variable $\xi$, then
$$
\sum_{n=0}^\infty a_n\mathbf 1_{\xi\geqslant n}=\sum_{n\leqslant\xi}a_n=\sum_{n=0}^{\lfloor\xi\rfloor}a_n,
$$
hence
$\sum\limits_na_nP[\xi\geqslant n]$ converges if and only if $A(\xi)$ is integrable, where $A(x)=\sum\limits_{k=1}^{\lfloor x\rfloor}a_k$. The case $a_n=n$ and $\xi=|X-E[X]|/k$ is used in the answer above, then $A(x)=\frac12\lfloor x\rfloor(\lfloor x\rfloor+1)$ hence $A(\xi)$ is integrable if and only if $\xi^2$ is.
A: If $X_0$ has a finite variance it is true. Using the inequality $(1-t)^N\geqslant 1-tN$ for $0\leqslant t\leqslant 1$ and independence, we are reduced to prove that the series $\sum_j c_j$ is convergent, with 
$$c_j:= j\cdot\mu\{|X_0-\mathbb E(X_0)|\geqslant kj\}.$$
Define $A_j:=\{kj\leqslant |X_0-\mathbb E(X_0)|\lt (j+1)k\}$. Then 
$$\sum_{n=1}^\infty c_n=\sum_{n=1}^\infty n\sum_{j\geqslant n}\mu(A_j)\leqslant \sum_j j^2\mu(A_j)\leqslant \frac 1{k^2}\sum_j\int_{A_j}|X_0-\mathbb E(X_0)|^2\mathrm d\mu=\frac1{k^2}\operatorname{Var}(X_0).$$ 
