Proof of Theorem <xref ref-type="statement" rid="thm1">2.1</xref>.
Without loss of generality, assume that ank>0 for k≤n, n≥1. Let q>0 such that (1+(1/αp))/2<q<1. For all i≤n, let
(2.5)Yni=-ani-1nα(q-1)I(aniXi<-nα(q-1))+XiI(ani|Xi|≤nα(q-1))+ani-1nα(q-1)I(aniXi>nα(q-1)),Unk=∑i=1kaniYni.
Write
(2.6)An=⋃j=1n(|anjXj|≥ɛ),Bn=⋃1≤i<j≤n((aniXi>nα(q-1),anjXj>nα(q-1))⋃(aniXi<-nα(q-1),anjXj<-nα(q-1))).

Firstly, we prove that
(2.7)(max1≤k≤n|Snk|<6ɛ)⊇Anc⋂(max1≤k≤n|Unk|<2ɛ)⋂Bnc=⋂j=1n(|anjXj|<ɛ)⋂(max1≤k≤n|Unk|<2ɛ)⋂⋂1≤i<j≤n[((aniXi≤nα(q-1))⋃(anjXj≤nα(q-1)))⋂1≤i<j≤n=⋂((aniXi≥-nα(q-1))⋃(anjXj≥-nα(q-1)))]=^Dn.

For any ω∈Dn, we have
(2.8)|anjXj|<ɛ, |anjYnj|≤|anjXj|<ɛ, ∀1≤j≤n, max1≤k≤n|Unk|<2ɛ,
and for any 1≤i<j≤n,
(2.9)aniXi≤nα(q-1), or anjXj≤nα(q-1),aniXi≥-nα(q-1), or anjXj≥-nα(q-1).
Hence
(2.10)a=^♯{i;1≤i≤n,aniXi(ω)>nα(q-1)}≤1,b=^♯{i;1≤i≤n,aniXi(ω)<-nα(q-1)}≤1,
where the symbol ♯A denotes the number of elements in the set A.

When a=b=0, then |aniXi(ω)|≤nα(q-1) for any 1≤i≤n; thus, Yni(ω)=Xi(ω), and therefore by (2.8),
(2.11)max1≤k≤n|Snk|=max1≤k≤n|Unk|<2ɛ<6ɛ.

When a=1, b=0 (or a=0, b=1), then there exists only an i0: 1≤i0≤n such that ani0Xi0(ω)>nα(q-1) (or ani0Xi0(ω)<-nα(q-1)), the remaining j, |anjXnj(ω)|≤nα(q-1); thus, Xj(ω)=Ynj(ω). If 1≤k≤i0-1, then Snk(ω)=Unk(ω). If i0≤k≤n, then by (2.8),
(2.12)max1≤k≤n|Snk(ω)|=max1≤k≤n|∑1≤i≤k,i≠i0aniXi(ω)+ani0Xi0(ω)|=max1≤k≤n|∑i=1kaniYni(ω)-an0Yni0(ω)+ani0Xi0(ω)|≤max1≤k≤n|∑i=1kaniYni(ω)|+|ani0Yni0(ω)|+|ani0Xi0(ω)|<2ɛ+ɛ+ɛ<6ɛ.

When a=b=1, then there exist 1≤i1, i2≤n such that ani1Xi1(ω)>nα(q-1), ani2Xi2(ω)<-nα(q-1), the remaining j, |anjXj(ω)|≤nα(q-1); thus, Xj(ω)=Ynj(ω). Without loss of generality, assume that i1≤i2. If 1≤k≤i1-1, then Snk(ω)=Unk(ω); if i1≤k<i2, then by (2.8),
(2.13)max1≤k≤n|Snk(ω)|≤max1≤k≤n|Unk(ω)|+|ani1Yni1(ω)|+|ani1Xi1(ω)|<2ɛ+ɛ+ɛ<6ɛ.
If k≥i2, then by (2.8),
(2.14)max1≤k≤n|Snk(ω)|=max1≤k≤n|∑1≤i≤k,i≠i1,i2aniXi(ω)+ani1Xi1(ω)+ani2Xi2(ω)|≤max1≤k≤n|Unk(ω)|+|ani1Yni1(ω)|+|ani2Yni2(ω)|+|ani1Xi1(ω)|+|ani2Xi2(ω)|<6ɛ.
Hence, (2.7) holds, that is:
(2.15)(max1≤k≤n|Snk|≥6ɛ)⊆An⋃(max1≤k≤n|Unk|≥2ɛ)⋃Bn.
Therefore, in order to prove (2.3), we only need to prove that
(2.16)∑n=1∞nαp-2P(An)<∞,(2.17)∑n=1∞nαp-2P(Bn)<∞,(2.18)∑n=1∞nαp-2P(max1≤k≤n|Unk|≥2ɛ)<∞, ∀ɛ>0.
By (2.1), (2.2), Xi≺X, and αp>1,
(2.19)∑n=1∞nαp-2P(An)≤∑n=1∞nαp-2∑j=1nP(|anjXj|≥ɛ)≤∑n=1∞nαp-2∑j=1nP(|Xj|≥ɛanj-1≥ɛcnα)≪∑n=1∞nαp-1P(|X|≥ɛcnα)=∑n=1∞nαp-1∑j=n∞P(ɛcjα≤|X|<ɛc(j+1)α)=∑j=1∞∑n=1jnαp-1P(ɛcjα≤|X|<ɛc(j+1)α)≤∑j=1∞jαpP(ɛcjα≤|X|<ɛc(j+1)α)≪E|X|p<∞.
That is, (2.16) holds.

By Lemma 1.1(ii), Xi≺X, and the definition of q, αp(1-2q)<-1,
(2.20)∑n=1∞nαp-2P(Bn)≤∑n=1∞nαp-2∑1≤i<j≤n(P(aniXi>nα(q-1))P(anjXj>nα(q-1))∑n=1∞nαp-2∑n=1∞nαp-2+P(aniXi<-nα(q-1))P(anjXj<-nα(q-1)))≪∑n=1∞nαpP2(|X|>cnαq)≤∑n=1∞nαpn-2αpq(E|X|p)2≪∑n=1∞nαp(1-2q)<∞.
That is, (2.17) holds.

In order to prove (2.18), firstly, we prove that
(2.21)max1≤k≤n|EUnk|=max1≤k≤n|E∑i=1kaniYni|⟶0, n⟶∞.

(
i
)
When α≤1, then p>1/α≥1; from EXi=0 and the definition of q, we have q<1, αpq>αp+1-αpq=1+αp(1-q)>1:(2.22)max1≤k≤n|E∑i=1kaniYni| ≤∑i=1nani|EYni|=∑i=1nani|E(Xi-Yni)| ≤∑i=1nani{E|Xi+ani-1nα(q-1)|I(aniXi<-nα(q-1))+E|Xi-ani-1nα(q-1)|I(aniXi>nα(q-1))} ≪∑i=1naniE|Xi|I(|aniXi|>nα(q-1))≤∑i=1naniE|Xi|(ani|Xi|nα(q-1))p-1 =∑i=1nanipE|Xi|pnα(1-q)(p-1) ≪n-αp+1+αp-α-αpq+αq=n-(αpq-1)-α(1-q) ⟶0, n⟶∞.

(
ii
)
When α>1, and p≥1, then E|X|<∞ from (2.2), thus,
(2.23)max1≤k≤n|E∑i=1kaniYni|≤∑i=1naniE|X|≪n-α+1⟶0, n⟶∞.

(
iii
)
When α>1, and p<1, by -(αp-1)-α(1-q)(1-p)<0, and -α(1-q)-(αpq-1)<0, we get
(2.24)max1≤k≤n|E∑i=1kaniYni|≤∑i=1nani(E|Xi|I(ani|Xi|≤nα(q-1))+ani-1nα(q-1)P(|aniXi|>nα(q-1)))≤∑i=1nanipE|Xi|p|aniXi|1-pI(ani|Xi|≤nα(q-1))+∑i=1nnα(q-1)anipn-αp(q-1)E|Xi|p≪n-(αp-1)-α(1-p)(1-q)+n-α(1-q)-(αpq-1)⟶0.
Hence, (2.21) holds; that is, for any ɛ>0, we have max 1≤k≤n |EUnk|<ɛ for all sufficiently large n. Thus,
(2.25)P(max1≤j≤n|Unj|≥2ɛ)≤P(max1≤j≤n|Unj-EUnj|>ɛ).

Let Y~ni=Yni-EYni. Obviously, Yni is monotonic on Xi. By Lemma 1.1(iii), {Yni;n≥1,i≤n} is also a sequence of PNQD random variables with EY~ni=0, by Lemma 1.2 and -1-α(1-q)(2-p)<-1:
(2.26)∑n=1∞nαp-2P(max1≤j≤n|Unj-EUnj|>ɛ) ≪∑n=1∞nαp-2log2n∑j=1nEanj2Ynj2 ≪∑n=1∞nαp-2log2n∑j=1n(Eanj2Xj2I(anj|Xj|≤nα(q-1))+n2α(q-1)P(anj|Xj|>nα(q-1))) ≤∑n=1∞nαp-2log2n∑j=1n(E|anjXj|pnα(q-1)(2-p)+n2α(q-1)-αp(q-1)E|anjXj|p) ≪∑n=1∞(nαp-1-αp+α(q-1)(2-p)+n-1+αp-αpq+2αq-2α)log2n =2∑n=1∞n-1-α(1-q)(2-p)log2n <∞.
This completes the proof of Theorem 2.1.