# Divergence of a series

1. Oct 22, 2009

### zeebo17

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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!

2. Oct 22, 2009

### Bohrok

Might it help to write

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

3. Oct 22, 2009

### fantispug

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

4. Oct 22, 2009

### 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 $$- \infty$$?

5. Oct 22, 2009

### Bohrok

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

6. Oct 22, 2009

### zeebo17

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

7. Oct 22, 2009

### 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?

8. Oct 22, 2009

### zeebo17

Wow, tonight is not my night. haha

$$\sum_{n = 1}^\infty 1 = + \infty$$

Great! Thanks!

9. Oct 22, 2009

### Office_Shredder

Staff Emeritus
Where did this assumption that an goes to infinity come from?

10. Oct 22, 2009

### 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 an goes to infinity, no? If it doesn't go to infinity, shouldn't that make the series diverge "more"?

11. Oct 23, 2009

### Office_Shredder

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

12. Oct 23, 2009

### zeebo17

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

13. Oct 23, 2009

### 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, $$(a_n)^2 < a_n \Rightarrow (a_n)^2 + a_n < 2a_n$$.

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.