Let , where , and .
Prove that
if and only if .
I want to give the proof for those who were interested.
First of all, a very indirect proof of this result can be found in [1, Lemma 2].
References
[1] B.G. Zhang and J.S. Yu, Existence of positive solutions for neutral differential equations,
Science in China Series A 35 (1992), no. 11, 1306--1313. View PDF
I will here give a shorter proof for a more generalized form.
Lemma 1. Let be an increasing divergent sequence and .
Then
if and only if
provided that
.____________________________(*)
Proof. We shall consider two different cases.
- Let
.
In this case, the claim is true since both the integral and the sum diverge.- Let
.
If we define
for ,
then is decreasing on .
We compute that
where we have changed the order of integration in the last step.
Then an equivalent claim reads as follows.
if and only if .
Define by for .
Adopting the convention that the empty sum is , we get
for all ,
which proves that
and
diverge or converge together since by the assumption (*), we learn that
and
diverge or converge together.
This completes the proof.________________________________
Corollary 1. If in addition to the assumptions of Lemma 1, we have
.___________________________(**)
Then
if and only if .
Remark. In Corollary 1, if we let and for , we get
the result in the first post. Trivially, in this case (*) and (**) hold.