Here's a challenge problem I just made up:

Let be a sequence of positive real numbers such that . Show that

.

Printable View

- Mar 20th 2011, 04:01 PMBruno J.Divergence of a series
Here's a challenge problem I just made up:

Let be a sequence of positive real numbers such that . Show that

. - Mar 20th 2011, 05:16 PMPaulRS
Let .

We want to show that diverges as .

( I chose this notation to suggest the similarity with , if )

We assume that otherwise the series would diverge trivially.

Let , then

We observe that diverges since and now by the limit comparison test - both differences are always positive-, we get that diverges too.

**Notation:** - Mar 21st 2011, 11:54 AMBruno J.
Very nice Paul!

Here's my solution:

It's well known that if and converges , then converges also. Let . We have

. Now since , we see that diverges, which implies that diverges. - Mar 26th 2011, 10:57 PMbkarpuz
Your solution is same as what came to my mind when I saw the question.

For those who might want to see a reference for the relation between the sum and the product used by Bruno J, please see [Theorem 7.4.6, 1].

I quote the result for convenience.

**Lemma**. Let be a sequence of nonnegative numbers, then

and converge or diverge together.

**References**

[1] L.S. Hahn and B. Epstein,*Classical Complex Analysis*, Jones and Bartlett Publishers, Inc., London, 1996.