Proof
Suppose that the series 1,2,3 converge. We prove that
converges as follows. We apply the proposition
(
Kolmogorov inequality for
series 1
) to the r.v.
for
:
Since the series 3 converge, the RHS vanishes as
:
According to the proposition
(
Probability based
criteria for AS convergence
) this implies that the series
converge a.s.

Since the series 2 also converge, we conclude that the series
converge a.s.

Note that
and the series 1 converge. Hence, the
and
are equivalent sequences and we already established that the
converges a.s. Hence, by the proposition
(
Property of equivalent
sequences of r.v.
) the
converges a.s.

Suppose that the series
converge. We prove that 1,2,3 converge as follows. Since
converges the
cannot be greater then
for infinitely many
.
Hence,
By the proposition
(
Borel-Cantelli lemma, part
2
) that the series 1 converge a.s. Hence, the
and
are equivalent and, by proposition
(
Property of equivalent
sequences of r.v.
), the series
converge. Then the series 2 converge.

To prove that the series 3 converge we apply the proposition
(
Kolmogorov inequality for
series 2
) to the r.v.
for
:
If 3 diverges then the RHS above tends to 0 as
.
Thus, the tail of
is not bounded a.s. and such series cannot converge. The contradiction shows
that the series 3 must converge.