# Convergence/Divergence test

1. May 10, 2012

### dan38

1. The problem statement, all variables and given/known data
(Edit) SERIES of n--> infinity ( -1/n)

2. Relevant equations

3. The attempt at a solution
First tried ratio test, ended up with 1 so it was inconclusive
Then tried integral test, but realized it's not a positive function
Now I'm stuck =[

Last edited: May 10, 2012
2. May 10, 2012

### Staff: Mentor

Are you trying to decide whether this series converges/diverges?
$$\sum_{n = 1}^{\infty}\frac{-1}{n}$$

If so, that's not at all obvious from your problem statment.

3. May 10, 2012

### sharks

$$\lim_{n\to\infty} -\frac{1}{n}=-\lim_{n\to\infty} \frac{1}{n}=0$$
Since the limit has a fixed value, therefore it __________.

4. May 10, 2012

### dan38

sorry I meant the series

5. May 10, 2012

### Staff: Mentor

Since we've established that you're dealing with a series, what other tests do you know, besides the ratio test and integral test (both of which apply to series with pos. terms)?

6. May 11, 2012

### dan38

would p-series be applicable?

7. May 11, 2012

### Staff: Mentor

Sure, why not?

$$\frac{1}{n} = \frac{1}{n^1}$$

8. May 11, 2012

### Whovian

And why ... does the Integral Test break down for nonpositive functions? And even if it does, note that it converges iff the series of its negative converges, and we can Integral Test on that.

Another famous proof of this is found here: http://math.stackexchange.com/questions/5115/why-does-1-x-diverge

9. May 14, 2012

### dan38

so does that mean the p-series can be used regardless if it's positive or negative?

10. May 15, 2012

### vela

Staff Emeritus
The p-series is a type of series, not a test per se. If you have a series of the form
$$\sum_{n=1}^\infty \frac{1}{n^p},$$ you can tell if it will converges simply by looking at the value of p. This is similar to determining whether the geometric series
$$\sum_{n=0}^\infty ar^n$$ will converge simply by looking at the value of r.

The idea then is to express your original series in terms of a p-series and then deduce whether it will converge or diverge depending on whether the corresponding p-series converges or diverges.

11. May 15, 2012

### Karamata

You can also use that $\displaystyle\sum_{p ~\mbox{prime}}\frac{1}{p}$ diverges, or $\displaystyle\sum_{k=1}^n \frac{1}{k} = \log n +\gamma + \varepsilon_{n}$ (where $\gamma$ is the Euler–Mascheroni constant and $\varepsilon_n \sim \frac{1}{2n}$ which approaches 0 as $n$ goes to infinity), and then say, let $n \to \infty$ and you will see

12. May 15, 2012

### sharks

That just made things more complicated, unless the OP is familiar with that method/theorem.

I think the p-series suggestion is perfectly acceptable in this simple problem.

13. May 15, 2012

### Karamata

The problem was that in my country we don't call it p-series (i was thinking that p must be prime, obviously not, because i didn't know what that test called p-series were).

I just wanted to show in other ways (which may be harder for someone).

14. May 15, 2012

### LCKurtz

I can't help but wonder whether the OP mistyped his problem in the first place and meant the series$$\sum_{n=1}^\infty \frac {(-1)^n} n$$

15. May 15, 2012

### sharks

16. May 15, 2012

### LCKurtz

I know he meant the series. The question is whether he meant to have $(-1)^n$ instead of $-1$.

17. May 15, 2012

### HallsofIvy

Staff Emeritus
Yes, he said he meant the series
$$\sum \frac{-1}{n}$$

but that is very different from the series
$$\sum \frac{(-1)^n}{n}$$

One converges and the other diverges! dan38, which of those two series do you mean?

18. May 15, 2012

### dan38

yeah I meant the first one
so since 1/n diverges, -1/n must therefore diverge by p-series?

19. May 15, 2012

### Bohrok

Since you know Ʃ 1/n diverges, you can use a property of sums where you can move a constant out in front.
$\sum \frac{-1}{n} = -1\sum \frac{1}{n}$
and it diverges exactly the same way as Ʃ 1/n except that it's negative.