Necessary and sufficient conditions for $\sum_{k=1}^{k_n}X_{nk} \stackrel{P}\to \gamma$ and $\max P(|X_{nk}|\ge \epsilon) \to 0$ There is a series of independent random variables $\{X_{nk};k=1,2,\cdots,k_n,n=1,2,\cdots\}$.
I need proof $\sum_{k=1}^{k_n}X_{nk} \stackrel{P}\to \gamma$ and $\underset{n\to\infty}{\lim}\underset{1\le k \le k_n }{max} P(|X_{nk}|\ge \epsilon) = 0 \iff$
$$
(1) \sum_{k=1}^{k_n}\int_{|x|\ge \epsilon}dF_{nk}(x)\to 0 \\
(2) \sum_{k=1}^{k_n}\int_{|x|<\epsilon} dF_{nk}(x)\to \gamma \\
(3) \sum_{k=1}^{k_n}\{\int_{|x|<\epsilon}x^2dF_{nk}(x)-(\int_{|x|<\epsilon}xdF_{nk}(x))^2\}\to0
$$
I converted condition (1) into $\sum_{k=1}^{k_n}P(|x|\ge \epsilon)\to 0$，
and got $\underset{1\le k \le k_n }{max} P(|X_{nk}|\ge \epsilon) \to 0$ easily. But I don’t know how to use (2) and (3) to get the result that the sum of random variables converges in probability.
I tried to use the method of truncating random variables to get the proof, but it ended in failure.
 A: Condition (2) should actually read $$ \sum_{k=1}^{k_n}\int_{|x|<\epsilon}x dF_{nk}(x)\to \gamma,$$
otherwise summing the terms in (1) and (2) would give $k_n$.
With this new condition (2), one can get the wanted convergence in probability: for a fixed $\varepsilon$, let $Y_{n,k}=X_{n,k}\mathbf{1}\{\lvert X_{n,k}\rvert \geqslant \varepsilon\}$ and $Z_{n,k}=X_{n,k}\mathbf{1}\{\lvert X_{n,k}\rvert \lt \varepsilon\}$.
Then
$$
\sum_{k=1}^{k_n}X_{n,k} -\gamma=\sum_{k=1}^{k_n} Y_{n,k}+\sum_{k=1}^{k_n}\left(Z_{n,k}-\mathbb E\left[Z_{n,k}\right] \right)+\sum_{k=1}^{k_n}\mathbb E\left[Z_{n,k}\right]-\gamma
$$
and

*

*$\sum_{k=1}^{k_n} Y_{n,k}\to 0$ in probability because $\mathbb P\left(\sum_{k=1}^{k_n} Y_{n,k}\neq 0\right)\leqslant \sum_{k=1}^{k_n}\mathbb P\left(\lvert X_{n,k}\rvert \geqslant \varepsilon\right)$.


*$\sum_{k=1}^{k_n}\left(Z_{n,k}-\mathbb E\left[Z_{n,k}\right] \right)\to 0$ in $\mathbb L^2$, by independent and assumption (3).


*$\sum_{k=1}^{k_n}\mathbb E\left[Z_{n,k}\right]\to\gamma$ by (2).
