Proof of Divergence of Series with Non-Negative Real Numbers

[zeebo17]

Homework Statement

Suppose (a_n) is a sequence of non-negative real numbers such that the series {\sum_{n=1}}^\infty a_n diverges. Prove that the series {\sum_{n=1}}^\infty \frac{a_n}{1+a_n} must also diverge.

Homework Equations

The Attempt at a Solution

I was thinking about looking at l=limsup(a_n) and perhaps the requirements on it in the root test in order to see if that could tell me something about the l=limsup \left( \frac{a_n}{1+a_n} \right), but I haven't had much luck.

Any suggestions?
Thanks!

 

[Bohrok]

Might it help to write

\frac{a_n}{a_n + 1} = \frac{a_n + 1 - 1}{a_n + 1} = 1 - \frac{1}{a_n + 1}
?

 

[fantispug]

It's kind of a backdoor route, but I'd prove the contrapositive and use a simple comparison test.

 

[zeebo17]

Since a_n is a non negative sequence can we assume that it diverges to + \infty so then 1 - \frac{1}{a_n + 1} diverges to - \infty?

 

[Bohrok]

You're not correct with the limit of the fraction part, but that part isn't really important. What is \sum_{n = 1}^\infty 1?

 

[zeebo17]

Oo oops! It's the lim (1 - 0) = 1. But then wouldn't the sequence then converge to 1 rather than diverging?

 

[Bohrok]

It doesn't converge to 1.
\sum_{n = 1}^\infty 1 = ~?

What do you get when you add an infinite number of 1's?

 

[zeebo17]

Wow, tonight is not my night. haha

<br /> \sum_{n = 1}^\infty 1 = + \infty<br />

Great! Thanks!

 

[Office_Shredder]

Where did this assumption that a_n goes to infinity come from?

 

[Bohrok]

\sum_{n=1}^\infty \left(1 - \frac{1}{a_n + 1}\right) = \sum_{n=1}^\infty 1 - \sum_{n=1}^\infty \frac{1}{a_n + 1} should diverge whether or not a_n goes to infinity, no? If it doesn't go to infinity, shouldn't that make the series diverge "more"?

 

[Office_Shredder]

Except that if a_n is small, both series are wildly divergent and you can't make that split in the first place

 

[zeebo17]

hmmm. Is there another way I should approach this problem?

 

[JG89]

You can assume that \frac{a_n}{a_n + 1} goes to 0 for n going to infinity, because if it didn't then the series would diverge anyway. \frac{a_n}{1 + a_n} = 1 - \frac{1}{1 + a_n}. If \lim_{n \rightarrow \infty} 1 - \frac{1}{1 + a_n} = 0 then we see that a_n \rightarrow 0 as well.

For large enough n, (a_n)^2 &lt; a_n \Rightarrow (a_n)^2 + a_n &lt; 2a_n.

Play around with that inequality to get a lower bound for \frac{a_n}{1 + a_n} so that the series for that lower bound diverges.