Let , where , and .

Prove that

if and only if .

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- March 26th 2011, 01:43 PMbkarpuzDivergence of improper integrals
Let , where , and .

Prove that

if and only if . - April 2nd 2011, 12:01 AMbkarpuz
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. - Let