What is the Convergence/Divergence of the Series -1/n?

[dan38]

Homework Statement

(Edit) SERIES of n--> infinity ( -1/n)

Homework Equations

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 =[
Any hints/advice would be nice

 

[Mark44]

Homework Statement

lim of n--> infinity ( -1/n)

Homework Equations

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 =[
Any hints/advice would be nice

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 statement.

 

[DryRun]

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

 

[dan38]

sorry I meant the series

 

[Mark44]

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)?

 

[dan38]

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)?

would p-series be applicable?

 

[Mark44]

Sure, why not?

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

 

[Whovian]

Then tried integral test, but realized it's not a positive function

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

 

[dan38]

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

 

[vela]

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.

 

[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

sorry for bad English

 

[DryRun]

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

sorry for bad English

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.

 

[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).

sorry for bad English

 

[LCKurtz]

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

 

[DryRun]

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

sorry I meant the series

 

[LCKurtz]

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

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

 

[HallsofIvy]

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?

 

[dan38]

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

 

[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.