# Help understanding the conditions of the Alternating Series!

1. May 20, 2012

### dan38

1. The problem statement, all variables and given/known data
Okay from what I have learnt to prove that a series converges via the alternating test, you must prove the following conditions
2. Relevant equations
1) an > 0
2) lim an (n--> infinity) = 0
and
3) a(n+1) < an
3. The attempt at a solution
However recently I've been encountering questions in my textbook where the first condition isn't filled, but the others two are and the answer is that it is converging
Doing further research online, I've found that most people simply disregard the first condition as well, i.e.

So Would anyone be able to clarify why the first condition is seemingly irrelevant?

2. May 20, 2012

### sharks

Post those questions and your attempts. It's always easier to explain and find your mistakes when facing the questions themselves.

3. May 20, 2012

### dan38

ttp://www.wolframalpha.com/input/?i=-%28%28-1%29^n%29%2F%28n%2B1%29+convergence

Wouldnt you expect that to diverge since
an = -1/(n+1)
which isnt greater than 0 for all values of n

4. May 20, 2012

### sharks

Don't you mean: $a_n = \frac{(-1)^n}{(n+1)}$? That would be wrong.
Use LaTeX to write it or just take a screenshot or picture of the whole question and attach it to your post.

5. May 20, 2012

### dan38

no that would be series itself
an = -1/(n+1)
i.e. the co-efficient of -1^n
so how would that converge??

dont quite know how to use latex yet sorry

6. May 20, 2012

### Ray Vickson

You need to present your notation more carefully (as written, your statements look all wrong), but if I understand your question correctly, you are worried about some series being convergent event though some of the convergence test conditions you were taught do not apply. That is simply because the conditions you were taught are SUFFICIENT, not necessary. In other words, if such-and-such conditions hold we have a convergent series. This does not say anything at all about what happens when the conditions do NOT hold.

RGV

7. May 20, 2012

### dan38

so if it doesn't hold, do I use another test?

8. May 20, 2012

### Ray Vickson

Give a specific example. To try to speak in generalities at this stage is dangerous: it might leave you "knowing" things that are wrong.

RGV

9. May 20, 2012

### dan38

where my series = the negative of this

(-1)^n
--------
(n+1)

10. May 20, 2012

### dan38

sorry just had another quick question, if i had the negative series =

(-1)^n
--------
n^0.5

Could I use the comparison test to the modulus of this and then use the p-series to determine if the 1/root(n) converges/diverges?

11. May 20, 2012

### Ray Vickson

So (assuming we start at n = 0), one series is
$$1 -\frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots,$$
while the other is
$$-1 + \frac{1}{2}- \frac{1}{3} + \frac{1}{4} - \cdots.$$
Do you honestly and truly not see that convergence/divergence answers are exactly the same in the two cases?

RGV